Lesson: Light MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaGabaKqzafaeaaaaaaaaa8qacaWFtacaaa@3D14@  Reflection and Refraction

Question: 1

Which of the following can make a parallel beam of light when light from a point source is incident on it?

(a) Concave mirror as well as convex lens

(b) Convex mirror as well as concave lens

(c) Two plane mirrors placed at 90° to each other

(d) Concave mirror as well as concave lens

Solution

a

Question: 2

A 10-mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is:

(a) 30 cm

(b) 20 cm

(c) 40 cm

(d) 60 cm

Solution

b

Question: 3

Under which of the following conditions a concave mirror can form an image larger than the actual object?

(a) When the object is kept at a distance equal to its radius of curvature

(b) When object is kept at a distance less than its focal length

(c) When object is placed between the focus and centre of curvature

(d) When object is kept at a distance greater than its radius of curvature

Solution

c

Question: 4

In given below figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is:

(a) 3 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaada GcaaqaaiaaiodaaSqabaaakeaadaGcaaqaaiaaikdaaSqabaaaaaaa @37BF@

(b) 2 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaada GcaaqaaiaaikdaaSqabaaakeaadaGcaaqaaiaaiodaaSqabaaaaaaa @37BF@

(c) 1 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca aIXaaabaWaaOaaaeaacaaIYaaaleqaaaaaaaa@3798@

(d) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOqaaeaaca aIYaaaleaaaaaaaa@36CD@

Solution

a

Question: 5

A light ray enters from medium A to medium B as shown in the following figure. The refractive index of medium B relative to A will be

(a) Greater than unity

(b) less than unity

(c) Equal to unity

(d) Zero

Solution

b

Question: 6

Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the figure. Which of the following could be inside the box?

(a) A rectangular glass slab

(b) A convex lens

(c) A concave lens

(d) A prism

Solution

a

Question: 7

A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the given below figure. Which of the following could be inside the box?

(a) Concave lens

(b) Rectangular glass slab

(c) Prism

(d) Convex lens

Solution

d

Question: 8

Which of the following statements is true?

(a) A convex lens has 4 dioptre powers having a focal length 0.25 m

(b) A convex lens has MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaGabaKqzagaeaaaaaaaaa8qacaWFtacaaa@3D54@ 4 dioptre powers having a focal length 0.25 m

(c) A concave lens has 4 dioptre powers having a focal length 0.25 m

(d) A concave lens has MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaGabaKqzagaeaaaaaaaaa8qacaWFtacaaa@3D54@ 4 dioptre powers having a focal length 0.25 m

Solution

a

Question: 9

Magnification produced by a rear-view mirror fitted in vehicles:

(a) is less than one

(b) is more than one

(c) is equal to one

(d) Can be more than or less than one depending upon the position of the object in front of it.

Solution

a

Question: 10

Rays from Sun converge at a point 15 cm in front of a concave mirror. Where an object should be placed so that size of its image is equal to the size of the object?

(a) 15 cm in front of the mirror

(b) 30 cm in front of the mirror

(c) Between 15 cm and 30 cm in front of the mirror

(d) More than 30 cm in front of the mirror

Solution

b

Question: 11

A full-length image of a distant tall building can definitely be seen by using:

(a) A concave mirror

(b) A convex mirror

(c) A plane mirror

(d) Both concave as well as plane mirror

Solution

b

Question: 12

In torches, search lights and headlights of vehicles the bulb is placed:

(a) Between the pole and the focus of the reflector

(b) Very near to the focus of the reflector

(c) Between the focus and centre of curvature of the reflector

(d) At the centre of curvature of the reflector

Solution

b

Question: 13

The laws of reflection hold good for:

(a) Plane mirror only

(b) Concave mirror only

(c) Convex mirror only

(d) All mirrors irrespective of their shape

Solution

d

Question: 14

The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in the given figure. Which one of them is correct?

(a) A

(b) B

(c) C

(d) D

Solution

b

Question: 15

You are given water, mustard oil, glycerin and kerosene. In which of these media a ray of light incident obliquely at same angle would bend the most?

(a) Kerosene

(b) Water

(c) Mustard oil

(d) Glycerin

Solution

d

Question: 16

Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in figure X?

(a) Fig. A

(b) Fig. B

(c) Fig. C

(d) Fig. D

Solution

d

Question: 17

Which of the following ray diagrams is correct for the ray of light incident on a lens shown in the following figure?

(a) Fig. A

(b) Fig. B

(c) Fig. C

(d) Fig. D

Solution

a

Question: 18

A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.

(a) Plane, convex and concave

(b) Convex, concave and plane

(c) Concave, plane and convex

(d) Convex, plane and concave

Solution

c

Question: 19

In which of the following, the image of an object placed at infinity will be highly diminished and point sized?

(a) Concave mirror only

(b) Convex mirror only

(c) Convex lens only

(d) Concave mirror, convex mirror, concave lens and convex lens

Solution

d

Short answer Questions

Question: 20

Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.

(a) Object is placed between device and its focus, image formed is enlarged and behind it.

(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.

(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.

(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

Solution

(a) Concave mirror

(b) Convex lens

(c) Concave lens

(d) Convex mirror

Question: 21

Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

Solution

The refraction of light in a glass slab can be shown as given below.

On passing through a rectangular glass slab, a ray of light bends first at the air-glass interface and then again at the glass- air interface.

The angle of refraction by first interface is equal to the angle of incidence at the second interface, i.e. r1=r2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaeyiiIaTaamOCaiaaigdacqGH9aqp cqGHGic0caWGYbGaaGOmaaaa@434B@ .

The final emergent ray is always parallel to the incident ray and is displaced through a distance.

Thus e=i MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaeyiiIaTaamyzaiabg2da9iabgcIi qlaadMgaaaa@41BE@

Question: 22

A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine? Support your answer with reason.

Solution

No, the bending of light will be different in different mediums since each medium has a different refractive index. The amount of light refracted or bent depends on the refractive index of the second medium. Since the refractive index of kerosene or turpentine is more than that of water, the pencil will appear to be more bent.

Question: 23

How is the refractive index of a medium related to the speed of light? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media?

Solution

Refractive index of medium is the ratio of the speed of light in vacuum (or air) to the speed of light in the medium. So, the refractive index,

n= c v MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabg2 da9maalaaabaGaam4yaaqaaiaadAhaaaaaaa@39E2@

Let one of the medium be m 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabaGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaamyBa8aadaWgaaWcbaWdbiaaigda a8aabeaaaaa@3DA9@ . Its refractive index is n 1 = c v 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0xe9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa aaleaacaaIXaaabeaakiabg2da9maalaaabaGaam4yaaqaaiaadAha daWgaaWcbaGaaGymaaqabaaaaaaa@3B8C@

Let the other medium be m 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabaGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaamyBa8aadaWgaaWcbaWdbiaaikda a8aabeaaaaa@3DAA@ . Its refractive index is n 2 = c v 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0xe9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa aaleaacaaIYaaabeaakiabg2da9maalaaabaGaam4yaaqaaiaadAha daWgaaWcbaGaaGOmaaqabaaaaaaa@3B8E@

n 21 = v 1 v 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0xe9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyinIWLaam OBamaaBaaaleaacaaIYaGaaGymaaqabaGccqGH9aqpdaWcaaqaaiaa dAhadaWgaaWcbaGaaGymaaqabaaakeaacaWG2bWaaSbaaSqaaiaaik daaeqaaaaaaaa@3E8B@

Question: 24

Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.

Solution

Given that, the refractive index of diamond with respect to glass is,

n dg = v g v d =1.6 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0xe9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaaeaaca WGUbWaaSbaaSqaaiaadsgacaWGNbaabeaakiabg2da9maalaaabaGa amODamaaBaaaleaacaWGNbaabeaaaOqaaiaadAhadaWgaaWcbaGaam izaaqabaaaaOGaeyypa0JaaGymaiaac6cacaaI2aaaaaa@414D@  

The absolute refractive index of glass is,

n g = C v g =1.5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa aaleaacaWGNbaabeaakiabg2da9maalaaabaGaam4qaaqaaiaadAha daWgaaWcbaGaam4zaaqabaaaaOaeaaaaaaaaa8qacqGH9aqpcaaIXa GaaiOlaiaaiwdaaaa@3F58@

The absolute refractive index of diamond is,

n d = C v d MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0xe9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa aaleaacaWGKbaabeaakiabg2da9maalaaabaGaam4qaaqaaiaadAha daWgaaWcbaGaamizaaqabaaaaaaa@3BC8@

Therefore, v g v d × c v g = n d =1.6×1.5=2.40 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVC0xe9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca WG2bWaaSbaaSqaaiaadEgaaeqaaaGcbaGaamODamaaBaaaleaacaWG KbaabeaaaaGccqGHxdaTdaWcaaqaaiaadogaaeaacaWG2bWaaSbaaS qaaiaadEgaaeqaaaaakiabg2da9iaad6gadaWgaaWcbaGaamizaaqa baGccqGH9aqpcaaIXaGaaiOlaiaaiAdacqGHxdaTcaaIXaGaaiOlai aaiwdacqGH9aqpcaaIYaGaaiOlaiaaisdacaaIWaaaaa@4DB5@

Question: 25

A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?

Solution

The given statement is correct.

Magnified virtual image

For the statement to be correct, the object has to be placed within 20 cm from the lens in the first case.

Magnified real image

For the statement to be correct, the object has to be placed between 20 cm and 40 cm.

Question: 26

Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?

Solution

Sudha should move the screen towards the lens to obtain a clear image of the building because as the object distance increases, the image distance decreases.

The approximate focal length of this lens will be 15 cm.

Question: 27

How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?

Solution

The power of a lens is inversely proportional to its focal length.

P= 1 f MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh9frVeeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOabaeqabaaabaGaamiuaiaad2dadaWcaaqaaiaadgdaaeaa caWGMbaaaaaaaa@3EA0@  

The lens of focal length 20 cm will have more power than the lens of focal length of 40 cm.

Therefore, the lens having focal length of 20 cm will provide more convergence.

Question: 28

Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.

Solution

In two plane mirrors, the incident ray and reflected ray will always be parallel to each other for any angle of incidence when the two plane mirrors are placed at right angle to each other. The following diagram illustrates this property.

Question: 29

Draw a ray diagram showing the path of rays of light when it enters with oblique incidence:

(i) From air into water

(ii) From water into air

Solution

Long answer Questions

Question: 30

Draw ray diagrams showing the image formation by a concave mirror when an object is placed:

(a) Between pole and focus of the mirror

(b) Between focus and centre of curvature of the mirror

(c) At centre of curvature of the mirror

(d) A little beyond centre of curvature of the mirror

(e) At infinity

Solution

Question: 31

Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) Between optical centre and focus of the lens

(b) Between focus and twice the focal length of the lens

(c) At twice the focal length of the lens

(d) At infinity

(e) At the focus of the lens

Solution

Question: 32

Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

Solution

The Laws of Refraction

(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

(ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media.

Let us consider the refraction of light through the following glass slab.

Here AO is the incident ray. OB is the refracted ray and BC is the emergent ray.

As per the law of refraction, the incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence (here O), all lie in the same plane.

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for light of a given colour and for a given pair of media. This law is also known as Snell’s law of refraction and is given as:

sini sinr =constant(refractiveindex) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeWaaSaaaeaaciGGZbGaaiyAaiaac6ga caaMc8UaaGPaVlaadMgaaeaaciGGZbGaaiyAaiaac6gacaaMc8UaaG PaVlaadkhaaaGaeyypa0Jaae4yaiaab+gacaqGUbGaae4CaiGacsha caGGHbGaaiOBaiaabshacaaMc8UaaGPaVlaabIcacaqGYbGaaeyzai aabAgacaqGYbGaaeyyaiaabogacaqG0bGaaeyAaiaabAhacaqGLbGa aGPaVlaaykW7caqGPbGaaeOBaiaabsgacaqGLbGaaeiEaiaabMcaaa a@6775@

Question: 33

Draw ray diagrams showing the image formation by a concave lens when an object is placed

(a) At the focus of the lens

(b) Between focus and twice the focal length of the lens

(c) Beyond twice the focal length of the lens

Solution

Question: 34

Draw ray diagrams showing the image formation by a convex mirror when an object is placed

(a) At infinity

(b) At finite distance from the mirror

Solution

Question: 35

The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of80 cm and the lens?

Solution

The magnification,

m= υ u =3,  u= 80 3 =26.67 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOabaeqabaaeaaaaaaaaa8qacaWGTbGaeyypa0ZdaiabgkHi TmaalaaabaGaeqyXduhabaGaamyDaaaacqGH9aqpcqGHsislcaaIZa GaaiilaaqaeamapeGaeyO0H4TaaiiOaiaadwhacqGH9aqpdaWcaaqa aiaaiIdacaaIWaaabaGaeyOeI0IaaG4maaaacqGH9aqpcqGHsislca aIYaGaaGOnaiaac6cacaaI2aGaaG4naiaabccacaqGJbGaaeyBaaaa aa@5676@

The object should be placed at a distance of 26.67 cm from the lens.

The image is real and inverted. The lens is convex.

Question: 36

Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Solution

As image could be real or virtual, there are two possible situations.

Case I: In case of concave mirror. 

M= 1 3 f=20 cm m= 1 3 = v u v= u 3 1 f = 1 v + 1 u 1 20 = 3 u + 1 u = 4 u 1 20 = 4 u u=80 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOabaeqabaaeaaaaaaaaa8qacaWGnbGaeyypa0JaeyOeI0Ya aSaaaeaacaaIXaaabaGaaG4maaaaaeabadGaamOzaiabg2da9iaaco bicaaIYaGaaGimaiaabccacaqGJbGaaeyBaaqaeamacaWGTbGaeyyp a0JaeyOeI0YaaSaaaeaacaaIXaaabaGaaG4maaaacqGH9aqpdaWcaa qaaiabgkHiTiaadAhaaeaacaWG1baaaaqaeamacaWG2bGaeyypa0Za aSaaaeaacaWG1baabaGaaG4maaaaaeabadWaaSaaaeaacaaIXaaaba GaamOzaaaacqGH9aqpdaWcaaqaaiaaigdaaeaacaWG2baaaiabgUca RmaalaaabaGaaGymaaqaaiaadwhaaaaabqaWaiabgkHiTmaalaaaba GaaGymaaqaaiaaikdacaaIWaaaaiabg2da9maalaaabaGaaG4maaqa aiaadwhaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaamyDaaaacqGH9a qpdaWcaaqaaiaaisdaaeaacaWG1baaaaqaeamacqGHsisldaWcaaqa aiaaigdaaeaacaaIYaGaaGimaaaacqGH9aqpdaWcaaqaaiaaisdaae aacaWG1baaaaqaeamacaWG1bGaeyypa0JaeyOeI0IaaGioaiaaicda caqGGaGaae4yaiaab2gaaaaa@7795@

Case II: In case of a convex mirror

M=+ 1 3 f=+20 cm m=+ 1 3 = v u v= u 3 1 f = 1 v + 1 u 1 20 = 3 u + 1 u = 2 u MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOabaeqabaaeaaaaaaaaa8qacaWGnbGaeyypa0Jaey4kaSYa aSaaaeaacaaIXaaabaGaaG4maaaaaeabadGaamOzaiabg2da9iabgU caRiaaikdacaaIWaGaaeiiaiaabogacaqGTbaabqaWaiaad2gacqGH 9aqpcqGHRaWkdaWcaaqaaiaaigdaaeaacaaIZaaaaiabg2da9maala aabaGaeyOeI0IaamODaaqaaiaadwhaaaaabqaWaiaadAhacqGH9aqp daWcaaqaaiabgkHiTiaadwhaaeaacaaIZaaaaaqaeamadaWcaaqaai aaigdaaeaacaWGMbaaaiabg2da9maalaaabaGaaGymaaqaaiaadAha aaGaey4kaSYaaSaaaeaacaaIXaaabaGaamyDaaaaaeabadWaaSaaae aacaaIXaaabaGaaGOmaiaaicdaaaGaeyypa0ZaaSaaaeaacqGHsisl caaIZaaabaGaamyDaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaWG1b aaaiabg2da9maalaaabaGaeyOeI0IaaGOmaaqaaiaadwhaaaaaaaa@6B1A@  

Therefore, u=40 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWq VvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaiaaciWacmaadaGabiaaea GaauaaaOqaaabaaaaaaaaapeGaamyDaiabg2da9iaacobicaaI0aGa aGimaiaabccacaqGJbGaaeyBaaaa@4252@

Question: 37

Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaGabaKqzagaeaaaaaaaaa8qacaWFtacaaa@3D54@ 50 cm. What is the nature of the lens and its power used by each of them?

Solution

Power of a lens is the degree of its convergence or divergence. It is the reciprocal of its focal length in metres. Its unit is dioptre.

Case 1

The power of the lens in the first case,

P 1 = 1 f = 1 50 100 =2D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOabaeqabaaaqaaaaaaaaaWdbeaacaWGqbWdamaaBaaaleaa peGaamymaaWdaeqaaOWdbiabg2da9maalaaabaGaamymaaqaaiaadA gaaaGaeyypa0ZaaSaaaeaacaaIXaaabaWaaSaaaeaacaaI1aGaaGim aaqaaiaaigdacaaIWaGaaGimaaaaaaGaeyypa0JaaGOmaiaaykW7ca qGebaaaaa@49FE@

It is a convex lens.

Case 2

The power of the lens in the second case,

  P 2 = 1 f = 1 50 100 =2D MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOabaeqabaaaqaaaaaaaaaWdbeaacaWGqbWaaSbaaSqaaiaa dkdaaeqaaOGaeyypa0ZaaSaaaeaacaWGXaaabaGaamOzaaaacqGH9a qpdaWcaaqaaiaaigdaaeaadaWcaaqaaiabgkHiTiaaiwdacaaIWaaa baGaaGymaiaaicdacaaIWaaaaaaacqGH9aqpcqGHsislcaaIYaGaaG PaVlaabseaaaaa@4B9B@

It is a concave lens.

Question: 38

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under:

Position of candle =12.0cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWq VvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaiaaciWacmaadaGabiaaea GaauaaaOqaaabaaaaaaaaapeGaeyypa0JaaGymaiaaikdacaGGUaGa aGimaiaaykW7caqGJbGaaeyBaaaa@42F4@

Position of convex lens =50.0cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWq VvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaiaaciWacmaadaGabiaaea GaauaaaOqaaabaaaaaaaaapeGaeyypa0JaaGynaiaaicdacaGGUaGa aGimaiaaykW7caqGJbGaaeyBaaaa@42F6@

Position of the screen =88.0cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWq VvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaiaaciWacmaadaGabiaaea GaauaaaOqaaabaaaaaaaaapeGaeyypa0JaaGioaiaaiIdacaGGUaGa aGimaiaaykW7caqGJbGaaeyBaaaa@4301@

(i) What is the focal length of the convex lens?

(ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?

(iii) What will be the nature of the image formed if he further shifts the candle towards the lens?

(iv) Draw a ray diagram to show the formation of the image in case(iii) as said above.

Solution

(i) Object distance, u=( 5012 )=38 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaamyDaiabg2da9iabgkHiT8aadaqa daqaa8qacaaI1aGaaGimaiabgkHiTiaaigdacaaIYaaapaGaayjkai aawMcaa8qacqGH9aqpcqGHsislcaaIZaGaaGioaiaabccacaqGJbGa aeyBaaaa@4A20@

Image distance =( 8850 )cm=38 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaeyypa0ZdamaabmaabaWdbiaaiIda caaI4aGaeyOeI0IaaGynaiaaicdaa8aacaGLOaGaayzkaaWdbiaabo gacaqGTbGaeyypa0JaaG4maiaaiIdacaqGGaGaae4yaiaab2gaaaa@492F@

Therefore, 1 v 1 u = 1 f MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeWaaSaaaeaacaaIXaaabaGaamODaaaa cqGHsisldaWcaaqaaiaaigdaaeaacaWG1baaaiabg2da9maalaaaba GaaGymaaqaaiaadAgaaaaaaa@42D8@  (f is the focal length)

1 f = 1 19 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaeyO0H49aaSaaaeaacaaIXaaabaGa amOzaaaacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaaGyoaaaaca aMc8Uaae4yaiaab2gaaaa@4667@

f=19cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaeyO0H4TaamOzaiabg2da9iaaigda caaI5aGaaGPaVlaabogacaqGTbaaaa@44D1@  

(ii) Object distance on shifting is ( 5031 )=19 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabaGaciaacaqabeaada abauaaaOqaamaabmaabaaeaaaaaaaaa8qacaaI1aGaaGimaiabgkHi TiaaiodacaaIXaaapaGaayjkaiaawMcaa8qacqGH9aqpcaaIXaGaaG yoaaaa@43AC@  cm.

So, u=19 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbmLMB1H hicL2BSLMB11garmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqee0evGueE0jxyaibaieYlh91rFfeu0dXdh9vqqj =hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXd ar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaada abauaaaOqaaabaaaaaaaaapeGaamyDaiabg2da9iabgkHiTiaaigda caaI5aaaaa@400F@  cm.

Since the object is placed at the focus, the image will be formed at infinity.

(iii) When the student further shifts the candle towards the lens, the lens forms enlarged, virtual and erect image of the candle.

(iv) The ray diagram showing the formation of the image is given below.