Wave Optics Ques 7
7. The 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 $D=10 d$ ?
[2016]
(a) $I_0$
(b) $\frac{I_0}{4}$
(c) $\frac{3}{4} I_0$
(d) $\frac{I_0}{2}$
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Answer:
Correct Answer: 7.(d)
Solution:
- (d) Let $P$ is a point infront of one slit at which intensity is to be calculated. From figure,
Path difference $=S_2 P-S_1 P$
$=\sqrt{D^{2}+d^{2}}-D=D(1+\frac{1}{2} \frac{d^{2}}{D^{2}})-D$
$=D[1+\frac{d^{2}}{2 D^{2}}-1]=\frac{d^{2}}{2 D}$
$\Delta x=\frac{d^{2}}{2 \times 10 d}=\frac{d}{20}=\frac{5 \lambda}{20}=\frac{\lambda}{4}$
Phase difference,
$ \Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2} $
So, resultant intensity at the desired point ’ $p$ ’ is $I=I_0 \cos ^{2} \frac{\phi}{2}=I_0 \cos ^{2} \frac{\pi}{4}=\frac{I_0}{2}$
BETA
