Q. White light may be considered to be a mixture of waves with $\lambda$ ranging between $3900\ \mathring{\mathrm{A}}$ and $7800\ \mathring{\mathrm{A}}$. An oil film of thickness $10,000\ \mathring{\mathrm{A}}$ is examined normally by reflected light. If $\mu = 1.4$, then the film appears bright for

A) $4308\ \mathring{\mathrm{A}},\ 5091\ \mathring{\mathrm{A}},\ 6222\ \mathring{\mathrm{A}}$
B) $4000\ \mathring{\mathrm{A}},\ 5091\ \mathring{\mathrm{A}},\ 5600\ \mathring{\mathrm{A}}$
C) $4667\ \mathring{\mathrm{A}},\ 6222\ \mathring{\mathrm{A}},\ 7000\ \mathring{\mathrm{A}}$
D) $4000\ \mathring{\mathrm{A}},\ 4667\ \mathring{\mathrm{A}},\ 5600\ \mathring{\mathrm{A}},\ 7000\ \mathring{\mathrm{A}}$
Explanation

(a) The film appears bright when the path difference
$(2\mu\ t\cos{r)}$ is equal to odd multiple of $\frac{\lambda}{2}$
$i.e.\ 2\mu t\cos r = (2n - 1)\lambda/2$ where $n = 1,\ 2,\ 3\ldots$
$\therefore\lambda = \frac{4\mu t\cos r}{(2n - 1)}$
$= \frac{4 \times 1.4 \times 10,000 \times 10^{- 10} \times \cos 0}{(2n - 1)} = \frac{56000}{(2n - 1)}\mathring{\mathrm{A}}$
$\therefore\lambda = 56000\mathring{\mathrm{A}},\ 18666\mathring{\mathrm{A}},\ 11200\mathring{\mathrm{A}},\ 8000\mathring{\mathrm{A}},\ 6222\mathring{\mathrm{A}},\ 5091\mathring{\mathrm{A}},\ 4308\mathring{\mathrm{A}},\ 3733\mathring{\mathrm{A}}$
The wavelength which are not within specified range are to be refracted

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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