JEE Advanced 2014 · Paper 1 · Physics Q2

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PhysicsMultiple correct+3Question 2 of 60

A light source, which emits two wavelengths $\lambda_1=400$ nm and $\lambda_2=600$ nm, 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 number of fringes for them within a distance $y$ on one side of the central maximum are $m_1$ and $m_2$ , respectively, then

\beta_2>\beta_1

m_1>m_2

From the central maximum, 3rd maximum of

\lambda_2

overlaps with 5th minimum of

\lambda_1

The angular separation of fringes for

\lambda_1

is greater than

\lambda_2

Answer and worked solution

Official answer: A, B, C

Fringe width $\beta=\dfrac{\lambda D}d\propto\lambda$ .

(A) $\lambda_2>\lambda_1$ , so $\beta_2>\beta_1$ . True.
(B) In a fixed distance $y$ , $m=y/\beta$ , so the narrower fringes of $\lambda_1$ give $m_1>m_2$ . True.
(C) 3rd maximum of $\lambda_2$ : $y=\dfrac{3(600)D}d=\dfrac{1800D}d$ (nm units). 5th minimum of $\lambda_1$ : $y=\dfrac{4.5(400)D}d=\dfrac{1800D}d$ . They coincide. True.
(D) Angular separation $\lambda/d$ is smaller for $\lambda_1$ . False.

(A), (B), (C)

Topics:Interference and Young's double slit

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JEE Advanced 2014 · Paper 1, Physics question 2