Question

A double slit of separation $0.1\text{mm}$ is illuminated by white light. A coloured interference pattern is formed on a screen $100\text{cm}$ away. If a pinhole is located in this screen at a distance of $2\text{mm}$ from the central fringe, the wavelengths in the visible spectrum $\left(4000\dot{\text{A}}\text{to}7000\dot{\text{A}}\right)$ which will be absent in the light transmitted through the pinhole is

• A. $4000\dot{\text{A}}$
• B. $5000\dot{\text{A}}$
• C. $6000\dot{\text{A}}$
• D. $7000\dot{\text{A}}$

Answer 1

The correct option is A $4000\dot{\text{A}}$

When destructive interference of a particular wavelength occurs at the pinhole, that particular wavelength will be absent in transmitted light through the pinhole.

As, $\Delta x=\frac{yd}{D}$

For destructive interference at pinhole,
$\Delta x=(2n-1)\frac{\lambda }{2}$

$\Rightarrow (2n-1)\frac{\lambda }{2}=\frac{yd}{D}$

$\Rightarrow \lambda =\frac{2yd}{(2n-1)D}$

Subsituting, $n=1,3,\mathrm{5...}etc$, we get

$\lambda =2\left(\frac{yd}{D}\right),\frac{2}{3}\left(\frac{yd}{D}\right),\frac{2}{5}\left(\frac{yd}{D}\right)...$

Here, $\frac{yd}{D}=\frac{(2\times {10}^{-3})(0.1\times {10}^{-3})}{1.0}=2\times {10}^{-7}\text{m}$

$\Rightarrow \frac{yd}{D}=2000\dot{\text{A}}$

So, wavelengths that will be absent in transmitted light are,

$\lambda =4000\dot{\text{A}},2680\dot{\text{A}},1600\dot{\text{A}}...$

Hence, option $\left(A\right)$ is correct.