Physics Class 12 NCERT Solutions: Chapter 10 Wave Optics Part 5

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Double slit diffraction

Double Slit Diffraction

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Q: 16. In double-slit experiment using light of wavelength, the angular width of a fringe formed on a distant screen is. What is the spacing between the two slits?

Answer:

Wavelength of light used,

Angular width of fringe,

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

Therefore, the spacing between the slits is. .

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 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 partition wall in a room 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.

Q: 18. Two towers on top of two hills are 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?

Answer:

Distance between the towers,

Height of the line joining the hills.

Thus, the radial spread of the radio waves should not exceed.

Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:

Aperture can be taken as:

Fresnel’s distance is given by the relation,

Where,

Wavelength of radio waves

Therefore, the wavelength of the radio waves is.

Q: 19. A parallel beam of light of wavelength falls on a narrow slit and the resulting diffraction pattern is observed on a screen away. It is observed that the first minimum is at a distance of from the centre of the screen. Find the width of the slit.

Answer:

Wavelength of light beam,

Distance of the screen from the slit,

For first minima,

Distance between the slits

Distance of the first minimum from the centre of the screen can be obtained as:

It is related to the order of minima as:

Therefore, the width of the slits is.