A light source, which emits two wavelengths $\lambda_{1}=400 n m$ and $\lambda_{2}=600 n m$, 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

(A) $\beta_{2}>\beta_{1}$; (B) $m_{1}>m_{2}$; (C) From the central maximum, $3^{\text {rd }}$ maximum of $\lambda_{2}$ overlaps with $5^{\text {th }}$ minimum of $\lambda_{1}$

(D) The angular separation of fringes for $\lambda_{1}$ is greater than $\lambda_{2}$

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Wave Optics

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A light source, which emits two wavelengths

λ_{1} = 400 nm

and

λ_{2} = 600 nm

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

λ_{1}

and

λ_{2}

are

β_{1}

and

β_{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

(A) $β_{2} > β_{1}$ ✓ Correct
(B) $m_{1} > m_{2}$ ✓ Correct
(C)From the central maximum, $3^{rd}$ maximum of $λ_{2}$ overlaps with $5^{th}$ minimum of $λ_{1}$ ✓ Correct
(D)The angular separation of fringes for $λ_{1}$ is greater than $λ_{2}$

Answer: (A), (B), (C)

Solution

Fringe width

β = λ \frac{D}{d}

. Since

λ_{2} > λ_{1}

, we have

β_{2} > β_{1}

, so (A) is correct.

Number of fringes in distance y:

m = \frac{y}{β}

. Since

β_{2} > β_{1}

, we have

m_{1} > m_{2}

, so (B) is correct.

For overlap:

n_{2} λ_{2} = (m_{1} - \frac{1}{2}) λ_{1}

. With

n_{2} = 3

3 \times 600 = (m_{1} - \frac{1}{2}) \times 400

, giving

m_{1} = 4.5 + 0.5 = 5

, which is the 5th minimum. So (C) is correct.

Angular separation

θ \approx \frac{λ}{d}

. Since

λ_{1} < λ_{2}

, angular separation for

λ_{1}

is smaller, not greater. So (D) is wrong.

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Wave Optics

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