Question

The angular width of the central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000\dot{A}$. When the slit is illuminated by light of another wavelength$\left(\lambda \right)$, the angular width decreases by $30\%$. Calculate the wavelength $\left(\lambda \right)$ of this light in $\dot{A}$. The same decrease in the angular-width of central maximum is obtained when the original apparatus is immersed in a liquid. Find the refractive index $\mu$ of the liquid.

• A. $\lambda =4000\dot{A},\mu =1.729$
• B. $\lambda =4200\dot{A},\mu =1.729$
• C. $\lambda =4200\dot{A},\mu =1.429$
• D. $\lambda =4000\dot{A},\mu =1.429$

Answer 1

The correct option is C $\lambda =4200\dot{A},\mu =1.429$


The angular width of central maxima is given by,

${\theta }_w=2\theta =\frac{2\lambda }{a}$

When the light of wavelength ${\lambda }_1=6000\dot{A}$ is used,

${\theta }_{w1}=\frac{2{\lambda }_1}{a}---\left(1\right)$

Angular width of central maxima when light of wavelength $\lambda$ is used

${\theta }_{w2}=\frac{2\lambda }{a}---\left(2\right)$

Dividing, $\frac{\left(2\right)}{\left(1\right)}$

$\frac{{\theta }_{w2}}{{\theta }_{w1}}=\frac{\lambda }{{\lambda }_1}$

Given that,
$\Rightarrow {\theta }_{w2}=0.7{\theta }_{w1}$

$\Rightarrow 0.7=\frac{\lambda }{6000\dot{A}}$

$\Rightarrow \lambda =0.7\times 6000\dot{A}$

$\Rightarrow \lambda =4200\dot{A}$

When the apparatus is immersed in a liquid of refractive index $\mu$,

By Snell's law, ${\lambda }_m=\frac{\lambda }{\mu }$

Accordingly, angular fringe width when the apparatus is immersed in a liquid is given by

${\theta }_{wl}=\frac{2{\lambda }_m}{a}=\frac{2\lambda }{\mu a}$

$\Rightarrow {\theta }_{wl}=\frac{{\theta }_w}{\mu }$

Given that, ${\theta }_{wl}=0.7{\theta }_w$

$\Rightarrow 0.7=\frac{1}{\mu }$

$\Rightarrow \mu =1.429$

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

$\begin{array}{c}\text{Why this question?}\\ \text{Asked in IIT-JEE 1996.}\end{array}$