SATHEE: Optics Question 475

Optics Question 475

Question: In a YDSE light of wavelength $λ = 5000$

$\overset{o}{A}$ is used which emerges in phase from two slits a distance $d = 3 \times 10^{- 7} m$ apart. A transparent sheet of thickness $t = 1.5 \times 10^{- 7} m$ , is refractive index n = 1.17, is placed over one of the slits. Where does the central maxima of the interference now appear?

Options:

A) $\frac{D (μ - 1) t}{2 d}$

B) $\frac{2 D (μ - 1) t}{d}$

C) $\frac{D (μ + 1) t}{d}$

D) $\frac{D (μ - 1) t}{d}$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] The path difference introduced due to introduction of transparent sheet is given by

$Δ x = (m - 1) t .$

If the central maxima occupies position of nth fringe, then $(μ - 1) t = n λ = d s i n θ$

$\sin θ = \frac{(μ - 1) t}{d} = \frac{(1.17 - 1) \times 1.5 \times 10^{- 7}}{3 \times 10^{- 7}} = 0.085$

Hence the angular position of central maxima is $θ = \sin^{- 1} (0.085) = 4.88^{\circ}$

For small angles $\sin θ ≃ θ ≃ \tan θ$

$\tan θ = \frac{y}{D}$

$\frac{y}{D} = \frac{(μ - 1) t}{d}$ Shift of central maxima is $y = \frac{D (μ - 1) t}{d}$

This formula can be used if D is given.

Optics Question 475

Question: In a YDSE light of wavelength $λ = 5000$

Options:

Answer:

Solution:

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Optics Question 475

Question: In a YDSE light of wavelength λ=5000

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: In a YDSE light of wavelength $λ = 5000$