Question

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distances from the slits. If the screen is moved through a distance of $5\times {10}^{-2}\text{m}$ towards the slits, the change in fringe width is $3\times {10}^{-5}\text{m}$. If the distance between the slits is ${10}^{-3}\text{m}$, calculate the wavelength of light used.

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

Answer 1

The correct option is A $6000\mathring{A}$

Given:
$d={10}^{-3}\text{m}$
$\Delta D=5\times {10}^{-2}\text{m}$
$\Delta \beta =3\times {10}^{-5}\text{m}$

Now, ${\beta }_1=\frac{\lambda D_1}{d}$

And, ${\beta }_2=\frac{\lambda D_2}{d}$

As the screen is moved towards the slits, $D_2<D_1\Rightarrow {\beta }_2<{\beta }_1$

${\beta }_1-{\beta }_2=\frac{\lambda }{d}(D_1-D_2)$
$\Rightarrow \Delta \beta =\frac{\lambda }{d}\left(\Delta D\right)$

$3\times {10}^{-5}=\frac{\lambda }{{10}^{-3}}\times 5\times {10}^{-2}$

$\therefore \lambda =6\times {10}^{-7}\text{m}=6000\mathring{A}$

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