Consider the optical system shown in Fig. The point source of light $S$ is having wavelength equal to $\lambda$. The light is reaching screen only after reflection. For point $P$ to besecond maxima, the value of $\lambda$ would be $(\mathrm{D}>>d$ and $d>>\lambda)$

• Correct answer
  A

  $\frac{12 d^{2}}{\mathrm{D}}$
• B

  $\frac{6 d^{2}}{\mathrm{D}}$
• C

  $\frac{3 d^{2}}{\mathrm{D}}$
• D

  $\frac{24 d^{2}}{\mathrm{D}}$

Solution

$\text { At } \mathrm{P}, \Delta x=\frac{(8 d) \times 3 d}{\mathrm{D}}$ For 2 nd maxima, $\Delta x=2 \lambda$ $\begin{array}{l} \Rightarrow \frac{24 d^{2}}{\mathrm{D}}=2 \lambda \\ \Rightarrow \lambda=\frac{12 d^{2}}{\mathrm{D}} \end{array}$

Hence, the correct answer is option: 'A'.