Question

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9 : 4 then what is the value of x?

(Assume that the field strength varies according to the slit width.)

(1) 2

(2) 3

(3) 5

(4) 4

Answer 1

Correct option is : (3) 5

\(I\propto(\text { width })^{2}\) 

\( \left(\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}=\frac{9}{4}\) 

\( \frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}=\frac{3}{2}\) 

\( \frac{(x+1) \mathrm{d}}{(\mathrm{x}-1) \mathrm{d}}=\frac{3}{2}\) 

\( \Rightarrow 3 \mathrm{x}-3=2 \mathrm{x}+2\) 

\( \mathrm{x}=5\)