Question

In Young's double slit experiment, intensity at a point is \( \dfrac{1}{4} \) of the maximum intensity. Angular position of this point is (separation between slits is \( d \)):

• A. \( \sin^{-1} \left( \dfrac{\lambda}{d} \right) \)
• B. \( \sin^{-1} \left( \dfrac{\lambda}{2d} \right) \)
• C. \( \sin^{-1} \left( \dfrac{\lambda}{3d} \right) \)
• D. \( \sin^{-1} \left( \dfrac{\lambda}{4d} \right) \)

Hint:

In Young's double slit experiment, the angular position of a point where the intensity is a fraction of the maximum can be found using the cosine or sine relationship involving the wavelength and slit separation.

Answer 1

The Correct Option is B

Step 1: In Young's double slit experiment, the intensity is given by the formula: \[ I = I_{\text{max}} \cos^2 \left( \dfrac{\pi d \sin \theta}{\lambda} \right). \] Step 2: For the intensity to be \( \dfrac{1}{4} \) of the maximum intensity, we have: \[ \cos^2 \left( \dfrac{\pi d \sin \theta}{\lambda} \right) = \dfrac{1}{4} \quad \Rightarrow \quad \dfrac{\pi d \sin \theta}{\lambda} = \dfrac{\pi}{3}. \] Step 3: Therefore, \[ \sin \theta = \dfrac{\lambda}{2d}. \]
Final Answer: \[ \boxed{\sin^{-1} \left( \dfrac{\lambda}{2d} \right)} \]