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Wave Optics — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsWave Optics1 Mark
MCQ
The maximum intensity in Young's double slit experiment is $l$. Distance between the slits is $d=5 \lambda$, where $\lambda$ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D=10 d$
  • A
    (a) $\frac{I_0}{2}$
  • ✓
    (b) $\frac{3}{4} I _0$
  • C
    (c) $I$
  • D
    (d) $\frac{I_0}{4}$
✓

Answer

Correct option: B.
(b) $\frac{3}{4} I _0$
(b) For maximum intensity on the screen$d \sin \theta=n \lambda \Rightarrow \sin \theta=\frac{n \lambda}{d}=\frac{n(2000)}{7000}=\frac{n}{3.5}$Since maximum value of $\sin \theta$ is 1So $n=0,1,2,3$, only. Thus only seven maximas can be obtained on both sides of the screen.

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