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DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET PDF Download

Introductory Exercise 29.1

Ques 1: In YDSE, D = 1.2 m and d = 0.25 cm. The slits are illuminated with coherent 600 nm light. Calculate the distance y above the central maximum for which the average intensity on the screen is 75% of the maximum.
Sol:

DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 48 × 10-6 m = 48 μm

Ques 2: A possible means for making an airplane invisible to radar is to coat the plane wit h an ant i reflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is μ =1.5. How thick is the oil film? Refractive index of the material of airplane wings is greater than the refractive index of polymer.
Sol: 
2μt = λ/2 (See the extra points)
This is the condition for destructive interference.
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

Ques 3: Slit 1 of a double slit is wider than slit 2, so that the light fro m slit 1 has an amplitude three times that of the light fro m slit 2. Show that Eq. (xii) in article 29.4 is replaced by the equation,
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
Sol:
A1 = 3A2
So, I1 = 9I2 
Let A2 = A0 and I2 = I0 
Then A1 = 3A0 and I1 = 9I0 
and Imax = 16 I0
Now, I = I1 + I2DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
Ques 4: Determine what happens to the double slit interference pattern if one of the slits is covered with a thin, transparent film whose thickness is DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET where λ is the wavelength of the incident light and μ is the index o f refraction o f the film.
Sol: 
Path difference produced by slab,DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET path difference is equivalent to 180° phase difference.
Hence, maximas and minimas interchange their positions.

Ques 5: Two slit s 4.0 × 10-6 m apart are illuminated by light of wavelength 600 nm. What is the highest order fringe in the interference pattern?
Sol:
Δx = d sin θ = nλ
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
Highest integer in Δ6.

Ques 6: Consider an interference experiment using eight equally spaced slit s. Determine the smallest phase difference in the waves from adjacent slits such that the resultant wave has zero amplitude.
Sol:

 DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

Exercises
For JEE Main

Subjective Questions
Energy Distribution in Interference


Ques 1: Two waves of the same frequency and same amplitude a are reaching a point simultaneously. What should be the phase difference between the waves so that the amplitude of the resultant wave be :
(i) 2a (ii) √2a (iii) a and (iv) zero?
Sol:
 DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

Ques 2: Two waves of equal frequencies have their amplitude in the ratio of 5 :3. They are superimposed on each other. Calculate the ratio of the maximum to minimum intensities of the resultant wave.
Sol:
Amax =5 + 3 = 8 units
Amin = 5 - 3 = 2 units
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

Ques 3: In a Young's double slit experiment λ = 500 nm, d = 1.0 mm and D = 1.0m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
Sol: 
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 1.25 × 10-4 m

Ques 4: In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength λ. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.
Sol:
(a) See the hint of above example. At half intensit y,
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
or  DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
∴  DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

Ques 5: In a two-slit interference pattern, the maximum intensity is I0.
(a) At a point in the pattern where the phase difference between the waves from the two slits is 60°, what is the intensity?
(b) What is the path difference for 480 nm light from the two slits at a point where the phase angle is 60°?
Sol: 
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
(b) 60° phase difference is equivalent to λ/6 path difference.

Conditions for Interference
Ques 6: Two coherent sources A and B of radio waves are 5.00 m apart . Each source emits waves with wavelength 6.00 m. Consider points along the line between the two sources. At what distances, if any, from A is the interference (a) constructive, (b) destructive?

Sol: (a) At centre path difference is zero. Therefore construction interference will be obtained.
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET At a distance, where path difference is λ/2 or 3m destructive interference will be obtained.
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
At P1 BP1 - AP1 = 3rn = λ/2
At P2 AP2 - BP2 = 3m = λ/2

Ques 7: A radio transmitting station operating at a frequency o f 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.00 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?
Sol:

DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
Δx = (BP - AP) = (9 - 2x) = nλ
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
Now, substituting n = 1, 2, ...... etc. We can find different values of x.
x1 = 3.25 m for n = 1
x= 2.0m for n = 2
and
x3 = 0.75 m for n = 3
Similarly we will get three points at same distance from other point B.

Young’s Double Slit Experiment
Ques 8: Coherent light from a sodium-vapour lamp is passed through a filter that blocks every thing except for light of a single wavelength. It then falls on two slits separated by 0.460 mm. In the resulting interference pattern on a screen 2.20 m away, adjacent bright fringes are separated by 2.82 mm. What is the wavelength?

Sol: DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 0.589 × 106m = 0.590 nm

Ques 9: Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is 2.0 x 10-3m.
Sol: 
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
≈ 0.014°

Ques 10: A Young's double slit apparatus has slits separated by 0.25 mm and a screen 48 cm away from the slit s. The whole apparatus is immersed in water and the slit s are illuminated by the red light ( λ = 700 nm in vacuum). Find the fringe-width of the pattern formed on the screen. (μw = 4/3)
Sol: 
Wavelength in water DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
Fringe width DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 10-3 m = 1 mm

Ques 11: In a two-slit experiment with monochromatic light, fringes are obtained on a screen place d at some distance from the slits. If the screen is moved by 1.5 × 10-2 m towards the slits, the change in fringe-width is 3 × 10-5 m. If the distance between the slits is 10-3 m, calculate the wavelength of the light used.
Sol: DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 2.0 × 10-6 m = 2.0 μm

Ques 12: In a double slit experiment the distance between the slits is 5.0 mm and the slit s are 1.0 m from the screen. Two interference patterns can be seen on the screen one due to light with wavelength 480 nm, and the other due to light with wavelength 600 nm. What is the separation on the screen between the third order bright fringes of the two interference patterns?
Sol:
DistanceDC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 7.2 × 10-5m = 0.072 mm

Ques 13: Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?
Sol: 
The required distance = one fringe width
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
= 8.33 × 10-4 m = 0.83 mm

Ques 14: Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m fro m the slit s. The first -order bright fringe is at 4.94 mm from the centre of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?
Sol: DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
⇒ λ2 = 2λ1 = 1200 nm

Ques 15: Two very narrow slits are spaced 1.80 μm apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light of λ = 550 nm? (Hint: The angle θ is not small).
Sol:
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET

 DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
∴ y1 = D tan θ1
= 35 tan 8.78° = 5.41 cm
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
DC Pandey Solutions: Interference and Diffraction of Light | Physics Class 12 - NEET
∴  θ2 = 27.27°
y2 = D tan θ2
= 35 tan 27.27°
= 18 cm
Δy = y2 - y= 12.6 cm

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FAQs on DC Pandey Solutions: Interference and Diffraction of Light - Physics Class 12 - NEET

1. What is interference of light?
Ans. Interference of light refers to the phenomenon where two or more light waves combine together and either reinforce or cancel each other out. This occurs when the waves meet at a particular point in space and undergo superposition. It leads to the formation of alternating bright and dark regions known as interference fringes.
2. How does interference of light occur?
Ans. Interference of light occurs when two or more coherent light waves, typically emitted from two sources or produced by the same source but with a slight phase difference, overlap and interfere with each other. This interference can be either constructive, resulting in brighter regions, or destructive, leading to darker regions.
3. What is the difference between interference and diffraction of light?
Ans. The main difference between interference and diffraction of light lies in the nature of the waves involved. Interference involves the superposition of two or more coherent waves, while diffraction refers to the bending or spreading of light waves as they pass through an aperture or encounter an obstacle. Interference produces distinct interference fringes, while diffraction leads to the bending of light waves around objects or through small openings.
4. What are the applications of interference and diffraction of light?
Ans. Interference and diffraction of light have numerous practical applications. They are used in various fields such as optics, telecommunications, holography, spectroscopy, and interferometry. Some specific applications include the creation of colorful patterns in soap bubbles, the production of interference filters for light manipulation, the construction of diffraction gratings for spectral analysis, and the development of optical storage devices like CDs and DVDs.
5. How can interference and diffraction be observed in everyday life?
Ans. Interference and diffraction phenomena can be observed in several everyday situations. For example, when sunlight passes through a narrow opening or between tree branches, it undergoes diffraction, resulting in the formation of bright and dark regions. Interference can be observed in thin films, where different colors are reflected due to constructive and destructive interference. Additionally, interference patterns can be seen when two ripples on the surface of water meet and interfere with each other.
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