## NCERT 12 Physics Wave Optics Chapter 10 Exercise

**Q.1. Monochromatic light of wavelength 589 nm incident from air on a water surface. What are the wavelength frequency and speed of (a) reflected and (b) refracted light? of water is 1.33.**

**Q.2. What is the shape of the wavefront in each of the following cases ?**

**(a) light diverging from point source.**

**(b) light emerging out of a convex lens when a point source is placed it its focus.**

**(c) the portion of the wavefront of light from a distant star intercepted by earth.**

Sol. (a) The geometrical shape of the wave front would

be diverging spherical wave front, as shown in.

(b) When a point source is placed at the focus of a convex lens,

the rays emerging form the lens are parallel. Therefore, the wave

front must be plane, as shown in.

(c) As the star (i.e. source of light) is very far off i.e. at infinity,

the wave front intercepted by earth must be a plane wave front

as shown in.

**Q.3. (a) The refractive index of glass is 1.5. What is the speed of light in glass ? (Speed of light in vacuum is 3×10 ^{8} ms^{-1}).**

(b) Is the speed of light in glass independent of colour of light ? If not, which of the two colours, red and violet travels slower in a glass prism ?

**Q.4. In Young’s double slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm, determine the wavelength of light used in the experiment.**

**Q.5. In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. What is the intensity of light at a point where path difference is λ/3 ?**

**Q.6. A beam of light consisting of two wavelengths 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) what is the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide ?**

**Q.7. In a double slit experiment, the angular width of a fringe is found to be 0.2**

^{0}on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in immersed in immersed in water ? Take refractive index of water to be 4/3.**Q.8. What is Brewster angle for air to glass transition?**

**Q.9. Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray?**

Sol. Here, λ = 5000 Å.

On reflection, there is no change in wavelength or frequency.

Wavelength of reflected light λ’ = λ = 5000Å.

**Q.10. Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.**

**Q.11. The 6563 Å H**

_{2 }line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the speed with which the star is receding from earth.Here, negative sign shows recession of star.

**Q.12. Explain how Newton’s corpuscular theory predicts the speed of light in a medium, say water, to be greater then the speed of light in vacuum. Is the prediction confirmed by the experimental determination of speed of light in water ? If not, which alternative picture of light is consistent with experiment?**

Sol. According to Newton’s corpuscular theory of light, corpuscles of light strike the interface XY, separating a denser medium from a rarer medium, the component of their velocity along XY remains the same :

If v_{1} is velocity of light in rarer medium (air)

V_{2} is velocity of light in denser medium (water)

I is angle of incidence, r is angle of refraction,

then component of v_{1} along XY = v_{2} sin r

,

**Q.13. You have learnt in the text how Huygens principle Leads to the laws of reflection and refraction. Us ethe same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the distance of the object from the mirror.**

**Sol.**P is a point object placed at a distance rfrom a plane mirror M

_{1 }M

_{2}. With P as centreand PO = r as radius, draw a spherical arc ; AB.This is the spherical wavefront from the object, Incident on M

_{1 }M

_{2}. If mirror were not present, the position of wave front AB would be A’ B’ where PP’ = 2 r. In the presence of the mirror, wave front AB would appear as A’’ PB’’ the figure, A’B’ where A’’ B’’ are two spherical arcs located symmetrically on either side of M

_{1}M

_{2}. A’’ PB’’. From simple geometry, we find OP = OP’, which was to be proved.

**Q 14. Let us list some of the factors which could possibly influence the speed of wave propagation :**

**(i) Nature of source (ii) direction of propagation (iii)motion of source and/or observer (iv) wave length (v) intensity of the wave.**

**On which of these factors, if any, does (a) the speed of light in vacuum (b) speed of light in a medium (say glass or water) depend ?**

Sol. (a) Speed of light in vacuum is an absolute, according to Einstein’s theory of relativity. It does not depend upon any of the factors listed above or any other factor.

(b) The speed of light in a medium like water or glass

(i) does not depend upon the nature of the source.

(ii) does not depend upon the direction of propagation, when the medium is isotropic.

(iii) does not depend upon motion of the source w.r.t the medium, but depends on motion of the observer relative to the medium.

(iv) depends on wavelength of light, being lesser for shorter wavelength and vice- versa.

(v) does not depend upon intensity of light.

**Q 15. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations :**

**(i) source at rest ; observer moving**

**(ii) source moving ; observer at rest.**

**The exact Doppler formulae for the case of light waves in vacuum, are however, strictly identical for these situations. Explain why this should be so. Would you expect formulae to be strictly identical for the two situations in case of light traveling in a medium ?**

Sol. Sound waves require a material medium for propagation. That is why situation (i) and (ii) are not identical physically though relative motion between the source and the observer is the same in the two cases. Infact, Doppler’s formulae for sound are different in the two cases.

For light waves traveling in vacuum, there is nothing to distinguish between the two situations. That is why the formulae are strictly identical.

For light propagating in a medium, situations (i) and (ii) are not identical. The formulae governing the two situations would obviously be different.

**Q 16. In double slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1**

^{0}. What is the spacing between the two slits?**Q.17. Answer the following questions :**

**(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band.**

**(b)In what way is diffraction from each slit related to interference pattern in a double slit experiment? (c) When a tiny circular obstacle is place in the path of light form a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?**

**(d) Two students are separated by a 7 m partition will in a room 10 m high. If both light and sound waves can bend round corners, how is it that the students are unable to see each other even though they can converse easily.**

**(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/ slits or around small obstacles)disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?**

Sol. (a) When width (a) of single slit is made double, the half angular width of central maximum which is λ/a, reduces to half. The intensity of central maximum will become 4 times. This is because area of central diffraction band would become 1/4th.

(b) If width of each slit is of the order of λ, then interference pattern in the double slit experiment is modified by the diffraction pattern from each of the two slits.

(c) This is because waves diffracted from the edges of circular obstacle interfere constructively at the centre of the shadow resulting in the formation of a bright spot.

(d) For diffraction of waves by obstacle/aperture, through a large angle, the size of obstacle/aperture should be comparable to

has definite value i.e. sound waves bend around the partition. Hence students can converse easily. (e) The ray optics assumption is used in understanding location and several other properties of images in optical instruments. This is because typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength of light. Therefore, diffraction or bending of waves is of no significance.

**Q 18. Two tower on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?**

Sol. In order that the hill may not obstruct the spreading radio beam, the radial spread of the beam over the hill 20 km away must not

**Q 19. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm form the centre of the screen. Find the width of the slit.**

**Q 20. Answer the following questions :**

**(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.**

**(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?**

Sol. (a) A low flying aircraft reflects the T.V. signal. The slight shaking on the T.V. screen may be due to interference between the direct signal and the reflected signal.

(b) Superposition principle follows from the linear character of the differential equation governing wavemotion. If y_{1 }and y_{2} are solutions of the wave equation, so is any linear combination of y_{1 }and y_{2} . When amplitudes are large (e.g. high intensity laser beams) and non- linear effects are important, the situation is far more complicated.

**Q 21. In deriving the single slit diffraction pattern, it was stated that intensity is zero at angles n λ/a. Justify this by suitably dividing the slit to bring out the cancellation.**

Therefore each of the smaller slits would send zero intensity in the direction of . Hence for the entire single slit, intensity at angle n λ/a would be zero.