Question

Q.8
In a YDSE apparatus, d = 2 mm, 2 = 600 nm, D = 1 m. The slits individually produce same intensity on the
screen. Find the position (distance from central maximum) of point where intensity is times of the
maximum intensity on screen.
(A) 0.01 mm
(B) 13 * 10+ mm (C) 0.05 mm
(D) 1.5 x 10 mm
answer is c
if any one got please share solution to me​

Answer 1

Answer:

C)0.05mm

Explanation:

We make use of the following formulas:-

I=I_max[cos^2(Ø/2)]----(1)

Ø/x=2pi/lambda----(2)

Refer the attachment for detailed solution!!

Hope this answer helped you