Wave Optics MCQs for NEET — Physics Questions with Answers

$$\begin{gathered} Let n_1 bright fringe from the first interference coincides with the n_2 \\ bright fringe of the second interference. \\ Then, n_1{\lambda }_1=n_2{\lambda }_2 \\ \frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}\Rightarrow \frac{n_1}{n_2}=\frac{700}{600}=\frac{7}{6} \\ So, 7th bright fringe of 1st will coincide with 6th bright fringe of 2nd. \\ Dis\tan ce from the central maxima of 1st=\frac{7\lambda D}{d}=\frac{7\times 600\times {10}^{-9}\times 0.8}{4\times {10}^{-4}}=8.4mm \end{gathered}$$

The angular width of the diffraction pattern is directly proportional to wavelength.

Transverse nature of light.

$$\begin{gathered} \left(1\right) \\ \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} = \frac{{\mathrm{a}}_1^2}{{\mathrm{a}}_2^2} = \frac{9}{1} \\ \frac{{\mathrm{I}}_{\max }}{{\mathrm{I}}_{\min }} = \frac{{\left({\mathrm{a}}_1 + {\mathrm{a}}_2\right)}^2}{{\left({\mathrm{a}}_1 - {\mathrm{a}}_2\right)}^2} = \frac{{\left(3 + 1\right)}^2}{{\left(3 - 1\right)}^2} = 4:1 \end{gathered}$$

Shift of fringe pattern = $\left(\mathrm{\mu }-1\right)\frac{tD}{d}$

$$\begin{gathered} \therefore \frac{30 \mathrm{D}\left(4800\times {10}^{-10}\right)}{\mathrm{d}}=\left(0.6\right)\mathrm{t}\frac{\mathrm{D}}{\mathrm{d}} \\ 30\times 4800\times {10}^{-10}=0.6 \mathrm{t} \\ \mathrm{t}=\frac{30\times 4800\times {10}^{-10}}{0.6}=\frac{1.44\times {10}^{-5}}{0.6}=24\times {10}^{-6} \end{gathered}$$

(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.

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

When unpolarized light is incident on the first polaroid P1, the transmitted light becomes linearly polarized. The intensity of this polarized light is reduced by half. When it passes through P3 which makes an angle of 45° with P1, the intensity is further reduced by cos^2(45°) = 1/2. Finally, when this light reaches P2, which is perpendicular to P1, the component parallel to the axis of P2 is blocked, reducing the intensity by another 1/2. Thus, the final intensity transmitted through P2 is (Io/2) * (1/2) * (1/2) = Io/8.

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

 $\therefore$Ratio of intensites is given by

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

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

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

(d) Ist minima is formed at a distance 

                 $Y=\frac{\lambda D}{a}$

For the distance of the first dark band of the diffraction pattern from the centre of the screen is given by position of Ist minima.

 i.e.         $Y=\frac{\lambda D}{a}$

where, $\lambda$=wavelength of parallel beams

          D= focal length

          a= width of linear aperture.

$\Rightarrow y=\frac{\left(5\times {10}^{-5}\right)\left(0.6\right)}{0.02\times {10}^{-2}}$   given

$Y=0.15 cm$

In Young's double-slit experiment, the intensity at the maxima on the screen is given by (2Io), where Io is the intensity from a single slit. When the distance between the screen and the slits is increased, the intensity from each slit decreases due to the spreading of the wavefront. The intensity from a single slit on the screen is inversely proportional to the square of the distance. Therefore, at a distance D = 10d, the intensity from each slit is (Io/10^2) = Io/100, and the total intensity on the screen is (2 * Io/100) = Io/50.