SATHEE: Optics 6 Question 3

Optics 6 Question 3

4. The figure shows a Young’s double slit experimental setup. It is observed that when a thin transparent sheet of thickness $t$ and refractive index $μ$ is put in front of one of the slits, the central maximum gets shifted by a distance equal to $n$ fringe widths. If the wavelength of light used is $λ, t$ will be

(a) $\frac{2 n D λ}{a (μ - 1)}$

(b) $\frac{2 D λ}{a (μ - 1)}$

(d) $\frac{n D λ}{a (μ - 1)}$

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Answer:

Correct Answer: 4. (*)

Solution:

Path difference introduced by a slab of thickness $t$ and refractive index $μ$ is given by

$Δ = (μ - 1) t$

Position of the fringe is $x = \frac{Δ D}{d} = \frac{(μ - 1) t D}{d}$

Also, fringe width is given by

$β = \frac{λ D}{d}$

According to the question,

$\begin{aligned} n β & = x \\ \Rightarrow & \frac{n λ D}{d} & = (μ - 1) t \frac{D}{d} \\ \Rightarrow n λ = (μ - 1) t & or t & = \frac{n λ}{(μ - 1)} \end{aligned}$

Alternate Solution

Path difference, $Δ = \frac{x d}{D} = (μ - 1) t$

$or t = \frac{x d}{(μ - 1) D}$

$\begin{aligned} Also, β = \frac{λ D}{d} \Rightarrow \frac{d}{D} = \frac{λ}{β} \\ ∴ t = \frac{x λ}{(μ - 1) β} = \frac{n λ D \times λ}{d (μ - 1) β} [∵ x = n β] \\ \Rightarrow t = \frac{n λ}{(μ - 1)} \end{aligned}$

So, no option is correct.

Optics 6 Question 3

Answer:

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Optics 6 Question 3

Answer:

Alternate 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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