The width of one of the two slits in Young's double slit experiment is d while that of the
other slit is $x d$ . 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.)

Question

The width of one of the two slits in Young's double slit experiment is d while that of the
other slit is

x d

. 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.)

Accepted Answer

Correct answer: B. (b) I ∝ ( width )^2 ( _1+ I_2 I_1- I_2 )^2= 94 _1+ _2 _1- I_2= 32 (x+1 )d (x-1 )d= 32 ⇒3x-3=2x+2x=5

Solution