Ncert Exercises & Solutions

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 center 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?

(a)
Hint: \(\beta =\frac{2D\lambda }{d}\)
Step 1: Find the size of the new central maxima.
\(\beta' =\frac{2D\lambda }{2d}=\frac{\beta }{2}\)
In a single slit diffraction experiment, if the width of the slit made double the original width, then the site of the central diffraction band gets reduced to half.
Step 2: Find the intensity of the central bands.
The intensity of the central diffraction band gets Increased up to four times. Because the area of central bands becomes 1/4th.

area of central diffraction band would become Read more on Sarthaks.com - https://www.sarthaks.com/1766024/answer-following-questions-single-diffraction-experiment-width-double-original-widtharea of central diffraction band would become Read more on Sarthaks.com - https://www.sarthaks.com/1766024/answer-following-questions-single-diffraction-experiment-width-double-original-width

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(b)
Hint: Consider the width of each slit of the same order.
The interference pattern in YDSE is modified by diffraction from each slot. The final pattern is the result of the interference of the diffracted wave from each slit.

(c)
Hint: waves diffracted from the edges of circular obstacles.
When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which Interferes constructively at Me center of the shadow. This constructive Interference produces a bright spot.

(d)
Hint: \(sin\theta =\frac{\lambda }{a}\)
Bending of waves by obstacles at 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)
Hint: Consider ray optics assumption.
The justification is that in ordinary optical instruments, the size of the aperture involved Is much larger than the wavelength of crone light used.

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