In a Young's double slit experiment slit separation 0.1mm, one observes a bright fringe at angle (1)/(40)rad by using light of wavelength λ _ 1 . When the light of wavelength λ _ 2 is used a bright fringe is seen at the

Correct answer: C. 625nm, 500nm dsinθ = nλ ∴ λ = (dsinθ)/(n) = 10 ^ - 4 × (1)/(40) × (1)/(n) ∴ λ = 2.5 × 10 ^ - 6 n = 2500 × 10 ^ - 9 n = (2500)/(n) (nm) Putting n = 1, λ _ 1 = 2500nm n = 2, λ _ 2 = 1250nm n = 3, λ _ 3 = 833nm n = 4, λ _ 4 = 625nm n = 5, λ _ 5 = 500nm n = 6, λ _ 6 = 357nm

JEE Main 201910 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Morning Shift
Subject: Physics
Chapter: Wave Optics

JEE Main 2019 · 10 January · Morning Shift · Physics · Wave Optics

In a Young's double slit experiment slit separation 0.1mm, one...

In a Young's double slit experiment slit separation

0.1mm,

one observes a bright fringe at angle

\dfrac{1}{40}rad

by using light of wavelength

{\lambda}_{1} .

When the light of wavelength

{\lambda}_{2}

is used a bright fringe is seen at the same angle in the same set up. Given that

{\lambda}_{1} and {\lambda}_{2}

are in visible range

\left(380nmto740nm\right),

their values are:

Options

A $400nm, 500nm$
B $380nm, 525nm$
C $625nm, 500nm$
D $380nm, 500nm$

Correct answer

C. $625nm, 500nm$

Step-by-step solution

dsin\theta = n\lambda

∴ \lambda = \dfrac{dsin\theta}{n} = {10}^{- 4} \times \dfrac{1}{40} \times \dfrac{1}{n}

∴ \lambda = \dfrac{2.5 \times {10}^{- 6}}{n} = \dfrac{2500 \times {10}^{- 9}}{n} = \dfrac{2500}{n}

\left(nm\right)

<br/><br/>Putting

n = 1, {\lambda}_{1} = 2500nm

<br/><br/>

n = 2, {\lambda}_{2} = 1250nm

<br/><br/>

n = 3, {\lambda}_{3} = 833nm

<br/><br/>

n = 4, {\lambda}_{4} = 625nm

<br/><br/>

n = 5, {\lambda}_{5} = 500nm

<br/><br/>

n = 6, {\lambda}_{6} = 357nm

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