Question

Angular width of central maximum in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000\mathring{A}$. When the slit is illuminated by light of another wavelength, then the angular width decreases by $30$%. The same decrease in angular width of central maximum is obtained when the original apparatus is immersed in a liquid. The refractive index of the liquid will be

• A. $1.25$
• B. $1.42$
• C. $1.67$
• D. $1.5$

Answer 1

Answer: (B)

$1.42$

Given, ${\lambda }_1=6000\mathring{A}$
${\lambda }_2=$?
Angular width of central maximum $=\frac{2\lambda }{a}$
${\theta }_1=\frac{2{\lambda }_1}{a}$ ...(i)
and ${\theta }_2=\frac{2{\lambda }_2}{a}$
$\because {\theta }_2={\theta }_1\times 0.7$
$\therefore {\theta }_1\times 0.7=\frac{2{\lambda }_2}{a}$ ....(ii)
From equations (i) and (ii), we get
$\frac{1}{0.7}=\frac{{\lambda }_1}{{\lambda }_2}$
$\Rightarrow {\lambda }_2={\lambda }_1\times 0.7=6000\times 0.7$
$=4200\mathring{A}$
When immersed in liquid,
${\lambda }_2=\frac{{\lambda }_1}{\mu }$
$\because {\theta }_2={\theta }_1\times 0.7$
$\Rightarrow \frac{2{\lambda }_2}{a}=\frac{2{\lambda }_1}{a}\times 0.7$ $[\because {\lambda }_2=\frac{{\lambda }_1}{\mu }]$
$\Rightarrow \frac{2{\lambda }_1/\mu }{a}=\frac{2{\lambda }_1}{a}\times 0.7$
$\Rightarrow \frac{1}{\mu }=0.7\Rightarrow \mu =1.42$