NEET Physics Wave Optics Questions Solved

(4)

$$\begin{gathered} Initial resolving power =\frac{2\mu Sin\theta }{1.22\lambda } (\mu Sin\theta =numerical aperture) \\ Final resolving power=\frac{{\displaystyle 2\times 2\mu Sin\theta }}{{\displaystyle 1.22\frac{\lambda }{2}}}=4\times \frac{{\displaystyle 2\mu Sin\theta }}{{\displaystyle 1.22\lambda }}=4\times initial resolving power \end{gathered}$$

(d) According to question, 5th fringe in air = 8th bright fringe in the medium 

        $\left(2\times 5-1\right)\frac{\lambda D}{2d}=8\frac{\lambda D}{\mu d}$

          $9\frac{\lambda D}{2d}=8\frac{\lambda D}{\mu d}$

$\Rightarrow \frac{9}{2}=\frac{8}{\mu }$

$\Rightarrow \mu =\frac{8\times 2}{9}$

$\therefore$ Refractive index of the medium.

          $\mu =\frac{16}{9}=1.7777=1.78$

(c) According to the question 

From the above diagram, Intensity transmitted through$P_3$ 

                       $I_2=\frac{I_o}{2}{\cos }^2{45}^{°}$

      $$\begin{gathered} \Rightarrow I_2=\frac{I_0}{2}\times {\left(\frac{1}{\sqrt{2}}\right)}^2 \\ \Rightarrow I_2=\frac{I_o}{4} \end{gathered}$$

Similarly, intensity transmitted through $P_2$

              $$\begin{gathered} I_3=\frac{I_0}{4}{\cos }^2{45}^{°} \\ \Rightarrow I_3\frac{I_o}{4}\times {\left(\frac{1}{\sqrt{2}}\right)}^2 \\ \Rightarrow I_3=\frac{I_o}{4}\times \frac{1}{2} \\ \Rightarrow I_3=\frac{I_o}{8} \end{gathered}$$

(b) It is given that $\frac{l_2}{l_1}=n\Rightarrow l_2=n{l}_1$

 $\therefore$Ratio of intensites is given by

 $\frac{l_{max}-l_{min}}{l_{max}+l_{min}}=\frac{{\displaystyle {\left(\sqrt{l_2}+\sqrt{l_1}\right)}^2\_{\left(\sqrt{l_2}-\sqrt{l_1}\right)}^2}}{{\left(\sqrt{l_1}+\sqrt{l_2}\right)}^2{\displaystyle +}{\displaystyle {\left(\sqrt{l_2}-\sqrt{l_1}\right)}^2}}$

          $=\frac{{\displaystyle {\left(\sqrt{{\displaystyle \frac{l_2}{l_1}}}{\displaystyle +}{\displaystyle 1}\right)}^2}{\displaystyle -}{\displaystyle {{\displaystyle \left(\sqrt{\frac{{\displaystyle l_2}}{{\displaystyle l_1}}-1}\right)}}^2}}{{\left(\sqrt{\frac{l_2}{l_1}}+1\right)}^2{\displaystyle +}{\displaystyle {{\displaystyle \left(\sqrt{\frac{{\displaystyle l_2}}{{\displaystyle l_1}}-1}\right)}}^2}}$

$=\frac{{\left(\sqrt{n}+1\right)}^2-{\left(\sqrt{n}-1\right)}^2}{{\left(\sqrt{n}+1\right)}^2+{\left(\sqrt{n}-1\right)}^2}=\frac{2\sqrt{n}}{n+1}$