Wave Optics MCQs with Answers – Part 5 (Class 12 Physics)

Explanation: \( \textbf{Given:} \) Fringe width in air is \(\beta_{\text{air}}=1.8\,\text{mm}\), and refractive index of liquid is \(\mu=1.5\). \( \textbf{Fringe width relation:} \) \[ \beta=\frac{\lambda D}{d} \] \( \textbf{Effect of liquid:} \) Frequency remains unchanged, but wavelength in the liquid becomes \[ \lambda_{\text{liq}}=\frac{\lambda_{\text{air}}}{\mu} \] \( \textbf{Since \(D\) and \(d\) are unchanged:} \) \[ \beta_{\text{liq}}=\frac{\beta_{\text{air}}}{\mu} \] \( \textbf{Substitution:} \) \[ \beta_{\text{liq}}=\frac{1.8\,\text{mm}}{1.5} \] \( \textbf{Calculation:} \) \[ \beta_{\text{liq}}=1.2\,\text{mm} \] \( \textbf{Final answer:} \) The new fringe width is \(1.20\,\text{mm}\). The reduction occurs because the wavelength in the liquid is smaller than in air.

Explanation: \( \textbf{Given:} \) \(\mu=1.5\), \(t=3.0\,\mu\text{m}\), and \(\lambda=600\,\text{nm}\). \( \textbf{Extra optical path due to sheet:} \) \[ \Delta x=(\mu-1)t \] \( \textbf{Substitution:} \) \[ \Delta x=(1.5-1)(3.0\,\mu\text{m}) \] \( \textbf{Calculation of extra path:} \) \[ \Delta x=0.5\times3.0\,\mu\text{m}=1.5\,\mu\text{m} \] \( \textbf{Fringe shift in number of fringes:} \) \[ N=\frac{\Delta x}{\lambda} \] \( \textbf{Unit conversion:} \) \[ 1.5\,\mu\text{m}=1500\,\text{nm} \] \( \textbf{Substitution:} \) \[ N=\frac{1500\,\text{nm}}{600\,\text{nm}}=2.5 \] \( \textbf{Final answer:} \) The central fringe shifts by \(2.5\) fringe widths. The values of \(D\) and \(d\) cancel when the shift is expressed in fringe-width units.

Explanation: \( \textbf{Given:} \) Fifth bright fringe position is \(y_5=10.0\,\text{mm}\), and pattern shift is \(s=4.0\,\text{mm}\). \( \textbf{Bright fringe position relation:} \) \[ y_n=n\beta \] \( \textbf{For the fifth bright fringe:} \) \[ 10.0\,\text{mm}=5\beta \] \( \textbf{Find fringe width:} \) \[ \beta=2.0\,\text{mm} \] \( \textbf{Shift in fringe-width units:} \) \[ N=\frac{s}{\beta} \] \( \textbf{Substitution:} \) \[ N=\frac{4.0\,\text{mm}}{2.0\,\text{mm}}=2 \] \( \textbf{Relation to extra optical path:} \) \[ N=\frac{\Delta x}{\lambda} \] \( \textbf{Therefore:} \) \[ \Delta x=2\lambda \] \( \textbf{Final answer:} \) The sheet introduces an extra optical path of \(2\lambda\).

Explanation: \( \textbf{Given:} \) \(I_1=I\), \(I_2=9I\), and \(\Delta\phi=60^\circ\). \( \textbf{Interference intensity formula:} \) \[ I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\Delta\phi \] \( \textbf{Square-root term:} \) \[ \sqrt{I_1I_2}=\sqrt{I\cdot9I}=3I \] \( \textbf{Cosine value:} \) \[ \cos60^\circ=\frac{1}{2} \] \( \textbf{Substitution:} \) \[ I_R=I+9I+2(3I)\left(\frac{1}{2}\right) \] \( \textbf{Calculation:} \) \[ I_R=10I+3I=13I \] \( \textbf{Final answer:} \) The resultant intensity is \(13I\). The interference term uses amplitude information through \(\sqrt{I_1I_2}\), not the direct product \(I_1I_2\).

Explanation: \( \textbf{Given:} \) \(I_{\max}=36I_0\) and \(I_{\min}=4I_0\). \( \textbf{Let the amplitudes be proportional to \(a_1\) and \(a_2\), with \(a_1\gt a_2\):} \) \[ I_{\max}\propto(a_1+a_2)^2 \] \[ I_{\min}\propto(a_1-a_2)^2 \] \( \textbf{Take square-root ratio:} \) \[ \frac{a_1+a_2}{a_1-a_2}=\sqrt{\frac{36I_0}{4I_0}}=3 \] \( \textbf{Solve:} \) \[ a_1+a_2=3a_1-3a_2 \] \[ 2a_1=4a_2 \] \[ \frac{a_1}{a_2}=2 \] \( \textbf{Intensity ratio:} \) \[ \frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=4 \] \( \textbf{Final answer:} \) The individual intensities are in the ratio \(4:1\).

Explanation: \( \textbf{Given:} \) \(\lambda=500\,\text{nm}\), \(D=2.0\,\text{m}\), \(d=0.25\,\text{mm}\), and \(y=14.0\,\text{mm}\). \( \textbf{Fringe width relation:} \) \[ \beta=\frac{\lambda D}{d} \] \( \textbf{Unit conversion:} \) \[ \lambda=5.0\times10^{-7}\,\text{m},\qquad d=2.5\times10^{-4}\,\text{m} \] \( \textbf{Substitution:} \) \[ \beta=\frac{(5.0\times10^{-7})(2.0)}{2.5\times10^{-4}}\,\text{m} \] \( \textbf{Calculation:} \) \[ \beta=4.0\times10^{-3}\,\text{m}=4.0\,\text{mm} \] \( \textbf{Position in fringe-width units:} \) \[ \frac{y}{\beta}=\frac{14.0}{4.0}=3.5 \] \( \textbf{Dark-fringe positions:} \) \[ y=\left(n+\frac{1}{2}\right)\beta \] \( \textbf{Comparison:} \) \[ 3.5=3+\frac{1}{2} \] \( \textbf{Counting from the central bright fringe:} \) \(n=0\) gives the first dark fringe, \(n=1\) gives the second, \(n=2\) gives the third, and \(n=3\) gives the fourth. \( \textbf{Final answer:} \) The point \(P\) is on the fourth dark fringe.

Explanation: \( \textbf{Given:} \) \(y=6.0\,\text{mm}\) and \(\beta=4.0\,\text{mm}\). \( \textbf{Position in fringe-width units:} \) \[ \frac{y}{\beta}=\frac{6.0}{4.0}=1.5 \] \( \textbf{Bright positions:} \) \[ y=n\beta \] \( \textbf{Dark positions:} \) \[ y=\left(n+\frac{1}{2}\right)\beta \] \( \textbf{Comparison:} \) \[ 1.5=1+\frac{1}{2} \] \( \textbf{Dark-fringe index:} \) \[ n=1 \] \( \textbf{Counting from the centre:} \) \(n=0\) is the first dark fringe, so \(n=1\) is the second dark fringe. \( \textbf{Final answer:} \) The point is the second dark fringe.

Explanation: \( \textbf{Given:} \) Fringe shift is \(s=3\beta\), and sheet thickness is \(t=4\lambda\). \( \textbf{Shift in fringe-width units:} \) \[ \frac{s}{\beta}=3 \] \( \textbf{Fringe shift due to sheet:} \) \[ \frac{s}{\beta}=\frac{(\mu-1)t}{\lambda} \] \( \textbf{Substitute \(t=4\lambda\):} \) \[ 3=\frac{(\mu-1)(4\lambda)}{\lambda} \] \( \textbf{Cancel \(\lambda\):} \) \[ 3=4(\mu-1) \] \( \textbf{Solve:} \) \[ \mu-1=\frac{3}{4} \] \[ \mu=1.75 \] \( \textbf{Final answer:} \) The refractive index of the sheet is \(1.75\).

Explanation: \( \textbf{Given:} \) Slit width \(a=0.40\,\text{mm}\), screen distance \(D=2.0\,\text{m}\), and central maximum width \(W=5.0\,\text{mm}\). \( \textbf{Central maximum width relation:} \) \[ W=\frac{2\lambda D}{a} \] \( \textbf{Solve for wavelength:} \) \[ \lambda=\frac{Wa}{2D} \] \( \textbf{Unit conversion:} \) \[ W=5.0\times10^{-3}\,\text{m},\qquad a=4.0\times10^{-4}\,\text{m} \] \( \textbf{Substitution:} \) \[ \lambda=\frac{(5.0\times10^{-3})(4.0\times10^{-4})}{2(2.0)}\,\text{m} \] \( \textbf{Numerator:} \) \[ (5.0\times10^{-3})(4.0\times10^{-4})=2.0\times10^{-6} \] \( \textbf{Division:} \) \[ \lambda=\frac{2.0\times10^{-6}}{4}=5.0\times10^{-7}\,\text{m} \] \( \textbf{Final answer:} \) The wavelength is \(500\,\text{nm}\).

Explanation: \( \textbf{Initial condition:} \) First minima are at \(y=\pm3.0\,\text{mm}\). \( \textbf{Initial central width:} \) \[ W_1=2(3.0\,\text{mm})=6.0\,\text{mm} \] \( \textbf{Single-slit width relation:} \) \[ W=\frac{2\lambda D}{a} \] \( \textbf{Change in screen distance:} \) \[ D_2=2D_1 \] \( \textbf{Change in slit width:} \) \[ a_2=\frac{a_1}{2} \] \( \textbf{Width ratio:} \) \[ \frac{W_2}{W_1}=\frac{D_2}{D_1}\cdot\frac{a_1}{a_2} \] \( \textbf{Substitution:} \) \[ \frac{W_2}{W_1}=2\times2=4 \] \( \textbf{Calculation:} \) \[ W_2=4(6.0\,\text{mm})=24.0\,\text{mm} \] \( \textbf{Final answer:} \) The new distance between the first minima is \(24.0\,\text{mm}\).

Explanation: \( \textbf{Given:} \) \(\theta_{\min}=7.5\times10^{-6}\,\text{rad}\), and \(\lambda=500\,\text{nm}\). \( \textbf{Rayleigh criterion for circular aperture:} \) \[ \theta_{\min}=\frac{1.22\lambda}{D_o} \] \( \textbf{Solve for objective diameter:} \) \[ D_o=\frac{1.22\lambda}{\theta_{\min}} \] \( \textbf{Unit conversion:} \) \[ \lambda=500\,\text{nm}=5.0\times10^{-7}\,\text{m} \] \( \textbf{Substitution:} \) \[ D_o=\frac{1.22(5.0\times10^{-7})}{7.5\times10^{-6}}\,\text{m} \] \( \textbf{Numerator:} \) \[ 1.22(5.0\times10^{-7})=6.10\times10^{-7} \] \( \textbf{Division:} \) \[ D_o=0.0813\,\text{m} \] \( \textbf{Convert to centimetres:} \) \[ 0.0813\,\text{m}=8.13\,\text{cm} \] \( \textbf{Final answer:} \) The required diameter is closest to \(8.1\,\text{cm}\).

Explanation: \( \textbf{Given:} \) \(\lambda=600\,\text{nm}\) and \(\text{NA}=0.75\). \( \textbf{Resolution relation:} \) \[ d_{\min}=\frac{0.61\lambda}{\text{NA}} \] \( \textbf{Convert wavelength:} \) \[ 600\,\text{nm}=0.600\,\mu\text{m} \] \( \textbf{Substitution:} \) \[ d_{\min}=\frac{0.61(0.600\,\mu\text{m})}{0.75} \] \( \textbf{Numerator:} \) \[ 0.61\times0.600=0.366 \] \( \textbf{Division:} \) \[ d_{\min}=\frac{0.366}{0.75}\,\mu\text{m} \] \( \textbf{Calculation:} \) \[ d_{\min}=0.488\,\mu\text{m} \] \( \textbf{Final answer:} \) The smallest resolvable distance is \(0.488\,\mu\text{m}\).

Explanation: \( \textbf{Resolution relation:} \) \[ d_{\min}\propto\frac{\lambda}{\text{NA}} \] \( \textbf{Old values:} \) \[ \lambda_1=600\,\text{nm},\qquad \text{NA}_1=0.60 \] \( \textbf{New values:} \) \[ \lambda_2=450\,\text{nm},\qquad \text{NA}_2=1.20 \] \( \textbf{Ratio:} \) \[ \frac{d_2}{d_1}=\frac{\lambda_2/\text{NA}_2}{\lambda_1/\text{NA}_1} \] \( \textbf{Substitution:} \) \[ \frac{d_2}{d_1}=\frac{450/1.20}{600/0.60} \] \( \textbf{Simplify using factors:} \) \[ \frac{d_2}{d_1}=\frac{450}{600}\times\frac{0.60}{1.20} \] \[ \frac{d_2}{d_1}=\frac{3}{4}\times\frac{1}{2}=\frac{3}{8} \] \( \textbf{Final answer:} \) The new limiting distance is \(\frac{3}{8}\) of the old one, so the resolving ability improves.

Explanation: \( \textbf{Incident intensity:} \) The light is plane-polarised with intensity \(I_0\). \( \textbf{First analyser angle from original direction:} \) \[ 30^\circ \] \( \textbf{After first analyser:} \) \[ I_1=I_0\cos^2 30^\circ=\frac{3I_0}{4} \] \( \textbf{Axis difference between first and second analysers:} \) \[ 60^\circ-30^\circ=30^\circ \] \( \textbf{After second analyser:} \) \[ I_2=I_1\cos^2 30^\circ=\frac{3I_0}{4}\times\frac{3}{4}=\frac{9I_0}{16} \] \( \textbf{Axis difference between second and third analysers:} \) \[ 90^\circ-60^\circ=30^\circ \] \( \textbf{After third analyser:} \) \[ I_3=I_2\cos^2 30^\circ=\frac{9I_0}{16}\times\frac{3}{4} \] \( \textbf{Final calculation:} \) \[ I_3=\frac{27I_0}{64} \] \( \textbf{Final answer:} \) The final transmitted intensity is \(\frac{27I_0}{64}\). Each transmitted beam becomes polarised along the latest analyser axis before the next projection is applied.

Explanation: \( \textbf{Start:} \) The incident light is plane-polarised with intensity \(I_0\). \( \textbf{Through the first analyser at \(45^\circ\):} \) \[ I_1=I_0\cos^2 45^\circ \] \[ I_1=I_0\times\frac{1}{2}=\frac{I_0}{2} \] \( \textbf{New vibration direction:} \) After the first analyser, the transmitted light is polarised along the first analyser axis. \( \textbf{Angle between the two analyser axes:} \) \[ 90^\circ-45^\circ=45^\circ \] \( \textbf{Through the second analyser:} \) \[ I_2=I_1\cos^2 45^\circ \] \[ I_2=\frac{I_0}{2}\times\frac{1}{2}=\frac{I_0}{4} \] \( \textbf{Final answer:} \) The final transmitted intensity is \(\frac{I_0}{4}\). The second projection uses the angle between successive analyser axes, not the angle between the original vibration and the final axis alone.

Explanation: \( \textbf{Incident light:} \) The light is unpolarised with intensity \(I\). \( \textbf{After first polariser:} \) \[ I_1=\frac{I}{2} \] \( \textbf{Angle between first and second axes:} \) \[ 30^\circ \] \( \textbf{After second polariser:} \) \[ I_2=I_1\cos^2 30^\circ \] \[ I_2=\frac{I}{2}\times\frac{3}{4}=\frac{3I}{8} \] \( \textbf{Angle between second and third axes:} \) \[ 30^\circ \] \( \textbf{After third polariser:} \) \[ I_3=I_2\cos^2 30^\circ \] \[ I_3=\frac{3I}{8}\times\frac{3}{4} \] \( \textbf{Final answer:} \) \[ I_3=\frac{9I}{32} \] The first polariser halves unpolarised light; the later factors come from Malus' law.

Explanation: \( \textbf{Given:} \) \(\tan i_p=1.5=\frac{3}{2}\). \( \textbf{Brewster condition:} \) \[ i_p+r=90^\circ \] \( \textbf{Therefore:} \) \[ r=90^\circ-i_p \] \( \textbf{Tangent relation for complementary angles:} \) \[ \tan(90^\circ-i_p)=\cot i_p \] \( \textbf{So:} \) \[ \tan r=\cot i_p=\frac{1}{\tan i_p} \] \( \textbf{Substitution:} \) \[ \tan r=\frac{1}{1.5}=\frac{2}{3} \] \( \textbf{Final answer:} \) The refracted angle satisfies \(\tan r=\frac{2}{3}\).

Explanation: \( \textbf{Given:} \) \(i_p=60^\circ\), and wavelength in air is \(\lambda_{\text{air}}=600\,\text{nm}\). \( \textbf{Brewster's law:} \) \[ \mu=\tan i_p \] \( \textbf{Find refractive index:} \) \[ \mu=\tan60^\circ=\sqrt{3} \] \( \textbf{Wavelength in medium:} \) \[ \lambda_{\text{med}}=\frac{\lambda_{\text{air}}}{\mu} \] \( \textbf{Substitution:} \) \[ \lambda_{\text{med}}=\frac{600\,\text{nm}}{\sqrt{3}} \] \( \textbf{Rationalise:} \) \[ \lambda_{\text{med}}=\frac{600\sqrt{3}}{3}\,\text{nm}=200\sqrt{3}\,\text{nm} \] \( \textbf{Final answer:} \) The wavelength inside the medium is \(200\sqrt{3}\,\text{nm}\), while the frequency remains unchanged.

Explanation: \( \textbf{Given:} \) \(\lambda=500\,\text{nm}=5.0\times10^{-7}\,\text{m}\), \(\mu=1.5\), and \(t=2.0\,\mu\text{m}=2.0\times10^{-6}\,\text{m}\). \( \textbf{Required:} \) Shift in terms of fringe width \(\beta\). \( \textbf{Extra optical path due to sheet:} \) \(\Delta x=(\mu-1)t\). \( \textbf{Substitution:} \) \(\Delta x=(1.5-1)(2.0\times10^{-6})\,\text{m}\). \( \textbf{Simplification:} \) \(\Delta x=0.5\times2.0\times10^{-6}=1.0\times10^{-6}\,\text{m}\). \( \textbf{Convert into wavelengths:} \) \(\frac{\Delta x}{\lambda}=\frac{1.0\times10^{-6}}{5.0\times10^{-7}}=2\). \( \textbf{Fringe-shift relation:} \) A path increase of \(2\lambda\) shifts the pattern by \(2\beta\). \( \textbf{Direction check:} \) The pattern shifts toward the slit in front of which the sheet is placed. \( \textbf{Final answer:} \) The fringe pattern shifts by \(2\beta\) toward the covered slit.

Explanation: \( \textbf{Given:} \) \(R=1.0\,\text{m}\), \(\lambda=500\,\text{nm}\), and \(n=4\). \( \textbf{Newton's ring relation for dark rings:} \) \[ r_n^2=n\lambda R \] \( \textbf{Convert wavelength:} \) \[ 500\,\text{nm}=5.0\times10^{-7}\,\text{m} \] \( \textbf{Substitution:} \) \[ r_4^2=4(5.0\times10^{-7})(1.0)\,\text{m}^2 \] \( \textbf{Calculation:} \) \[ r_4^2=2.0\times10^{-6}\,\text{m}^2 \] \( \textbf{Take square root:} \) \[ r_4=\sqrt{2.0\times10^{-6}}\,\text{m} \] \[ r_4=1.41\times10^{-3}\,\text{m} \] \( \textbf{Final answer:} \) The radius is \(1.41\,\text{mm}\).