Question

In a double slit experiment if light of wavelength $5000{\text{A}}^{\circ }$ is used then fringe width of $1\text{mm}$ is obtained. If now light of wavelength $6000{\text{A}}^{\circ }$ is used without altering the system then new fringe width will be :

• A. $1\text{mm}$
• B. $1.5\text{mm}$
• C. $0.5\text{mm}$
• D. $1.2\text{mm}$

Answer 1

The correct option is D $1.2\text{mm}$

As fringe width is given by , $\beta =\frac{\lambda D}{d}$

$1\times {10}^{-3}=(5000\times {10}^{-10})\times \frac{D}{d}$

$\Rightarrow \frac{D}{d}=\frac{\left({10}^{-3}\right)}{(5\times {10}^{-7})}$

When the light of wavelength $6000A^{\circ }$ is used then,

$\beta =6000\times {10}^{-10}\times \frac{\left({10}^{-3}\right)}{(5\times {10}^{-7})}$

$\Rightarrow \beta =1.2\text{mm}$

Hence, option (c) is the correct answer.