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JEE Advanced 2014Paper 1

Exam: JEE Advanced
Year: 2014
Paper: Paper 1
Subject: Physics
Chapter: Wave Optics

JEE Advanced 2014 · Paper 1 · Physics · Wave Optics

A light source, which emits two wavelengths λ _ 1 = 400nm and λ _ 2...

A light source, which emits two wavelengths

{\lambda}_{1} = 400nm

and

{\lambda}_{2} = 600nm

, is used in a Young's double slit experiment. If recorded fringe widths for

{\lambda}_{1}

and

{\lambda}_{2}

are

{\beta}_{1}

and

{\beta}_{2}

and the number of fringes for them within a distance y on one side of the central maximum are

{m}_{1}

and

{m}_{2}

, respectively, then

Options

A ${\beta}_{2} > {\beta}_{1}$
B ${m}_{1} > {m}_{2}$
CFrom the central maximum, ${3}^{rd}$ maximum of ${\lambda}_{2}$ overlaps with ${5}^{th}$ minimum of ${\lambda}_{1}$
DThe angular separation of fringes for ${\lambda}_{1}$ is greater than ${\lambda}_{2}$

Correct answer

A. ${\beta}_{2} > {\beta}_{1}$

Step-by-step solution

\beta = \dfrac{D\lambda}{d}

∵ {\lambda}_{2} > {\lambda}_{1} ⇒ {\beta}_{2} > {\beta}_{1}

Also

{m}_{1}{\beta}_{1} = {m}_{2}{\beta}_{2} ⇒ {m}_{1} > {m}_{2}

Also

3

\left(\dfrac{D}{d}\right)

\left(600nm\right) = \left(2 \times 5 - 1\right)\left(\dfrac{D}{2d}\right)400nm

Angular width

\theta = \dfrac{\lambda}{d}

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