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Page No 384: - Chapter 10 Wave Optics Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 10.11:

The 6563 Å https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7431/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_70fba3f6.gif  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 the Earth.

Answer:

Wavelength of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7431/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_70fba3f6.gif line emitted by hydrogen,

λ = 6563 Å

= 6563 × 10−10 m.

Star’s red-shift, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7431/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_m35e2a7cb.gif

Speed of light, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7431/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_m824fe28.gif

Let the velocity of the star receding away from the Earth be v.

The red shift is related with velocity as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7431/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_45961c56.gif

Therefore, the speed with which the star is receding away from the Earth is 6.87 × 105 m/s.

Question 10.12:

Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

Answer:

No; Wave theory

Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.

Hence, we can write the expression:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7432/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_561a9334.gif  … (i)

Where,

i = Angle of incidence

r = Angle of reflection

c = Velocity of light in air

v = Velocity of light in water

We have the relation for relative refractive index of water with respect to air as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7432/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_69d1b851.gif

Hence, equation (i) reduces to

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7432/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_3616bdd9.gif

But, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7432/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_m4ae39fa.gif > 1

Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of > v.

The wave picture of light is consistent with the experimental results.

Question 10.13:

You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the 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 object distance from the mirror.

Answer:

Let an object at O be placed in front of a plane mirror MO’ at a distance r (as shown in the given figure).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7433/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_27cd6780.jpg

A circle is drawn from the centre (O) such that it just touches the plane mirror at point O’. According to Huygens’ Principle, XY is the wavefront of incident light.

If the mirror is absent, then a similar wavefront X’Y’ (as XY) would form behind O’ at distance r (as shown in the given figure).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7433/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_1d98a1e.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7433/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_m5c3e698c.gif can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).

Question 10.14:

Let us list some of the factors, which could possibly influence the speed of wave propagation:

(i) Nature of the source.

(ii) Direction of propagation.

(iii) Motion of the 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) The speed of light in a medium (say, glass or water), depend?

Answer:

(a) Thespeed of light in a vacuum i.e., 3 × 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.

(b) Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.

Question 10.15:

For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

Answer:

No

Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.

In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

Question 10.16:

In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?

Answer:

Wavelength of light used, λ = 6000 nm = 600 × 10−9 m

Angular width of fringe,https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7437/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_333496bd.gif

Angular width of a fringe is related to slit spacing (d) as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7437/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_m37b5c60c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7437/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_3cda6059.gif

Therefore, the spacing between the slits ishttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/254/7437/NS_6-11-08_Sravana_12_Physics_10_21_NRJ_SG_html_5a189a3.gif .

Question 10.17:

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 the interference pattern in a double-slit experiment?

(c) When a tiny circular obstacle is placed in the path of light from 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 wall in a room 10 m high. If both light and sound waves can bend around obstacles, 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?

Answer:

(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.

(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.

(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.

On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.

(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.

Also Read : Page-No-385:-Chapter-10-Wave-Optics-Additional-Exercise-Solutions-class-12-ncert-solutions-Physics

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