NEET-XII-Physics
exam-1 year:2016
- #18The intensity at the maximum in a Young's double
slit experiment is ``I_0``. Distance between two slits is
d = 5 ``\lambda``, where ``\lambda`` is the wavelength of light used
in the experiment. What will be the intensity in front
of one of the slits on the screen placed at a distance
D = 10 d ?
(1) ``I_0``
(2)``I_0``/4
(3) 3/4 ``I_0``
(4)``I_0``/2digAnsr: 4Ans : 4
Sol. Path difference
= S2P - S1P
= + 2 2D d D
S1
S2
O B
P
D
d
= D
+
2
2
1 d
1 D
2 D
= D
+ =
2 2
2
d d
1 1
2D2D
▵x =
= = =
2d d 5
2 10d 20 20 4
▵ =
=
2
.
4 2
So, intensity at the desired point is
I = I0cos2
2
= I0cos2
4
= 0
I
2