Question

In Young's experiment $\lambda =$ 4000Å fringes observed have a width $\beta$ . The light illuminating with new $\lambda =$ 6000Å and the separation between the interfering sources is halved. The ratio of the distance between the screen and the interfering sources if the fringe width remains unaltered respectively is:

• A. 1/3
• B. 3/1
• C. 3/4
• D. 2/3

Answer 1

Answer: (B)

3/1

Given $:\frac{d_1}{d_2}=2$
fringe width $\left(\beta \right)=\frac{D\lambda }{d}$
So, ${\beta }_1=\frac{D_1{\lambda }_1}{d_1}$
and ${\beta }_2=\frac{D_2{\lambda }_2}{d_2}$
Now given ${\beta }_1={\beta }_2$
So, $\frac{D_1{\lambda }_1}{d_1}=\frac{D_2{\lambda }_2}{d_2}$
$\frac{D_1}{D_2}=\frac{{\lambda }_2{d}_1}{{\lambda }_1{d}_2}$
$=\left(\frac{{\lambda }_2}{{\lambda }_1}\right)\left(\frac{d_1}{d_2}\right)$
$=\left(\frac{6000}{4000}\right)\left(2\right)$
$=\frac{3}{1}$