17. The intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is d = 5λ, where λ 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 ?

17. The intensity at the maximum in a Young’s double slit experiment is I_0. Distance between two slits is d = 5λ, where λ 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 ?

\( (a)\; I_0 \)  

\( (b)\; \tfrac{I_0}{4} \)  

\( (c)\; \tfrac{3}{4} I_0 \)  

\( (d)\; \tfrac{I_0}{2} \)  

Answer :  \( (d)\; \tfrac{I_0}{2} \)  

\( \text{Path diff.} = S_{2}P - S_{1}P \)

\( = \sqrt{(10d)^{2} + d^{2}} - 10d \)

\( = d\left(\sqrt{101} - 10\right) \approx 0.05d \)

\( = 0.25\lambda = \tfrac{\lambda}{4} \)

\( \text{Phase diff. } \phi = \tfrac{2\pi}{\lambda} \times \text{Path diff.} \)

\( = \tfrac{2\pi}{\lambda} \cdot \tfrac{\lambda}{4} \)

\( = \tfrac{\pi}{2} \;\; \text{Minima} \)

\( I_{\text{max}} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos 0 \)

\( = 2I_{1} + 2I_{1} \)

\( = 4I_{1} = I_{0} \; (\text{given}) \)

\( I_{\text{min}} = I_{1} + I_{2} - 2\sqrt{I_{1}I_{2}}\cos 90^{\circ} \)

\( = 2I_{1} \)

\( = \tfrac{I_{0}}{2} \)