Wave Optics

${\lambda }_1=560nm,$ fringe width, B = 7.2 mm

$\Rightarrow B_1=\frac{D{\lambda }_1}{d}=7.2mm$

For ${\lambda }_2\to B_2=\frac{D{\lambda }_2}{d}=8.1mm$ ${}_{}{}_{}\frac{{}_{}}{}$

$\frac{B_2}{B_1}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{8.1}{7.2}\Rightarrow {\lambda }_2\frac{81}{72}{\lambda }_1$

$=\frac{9}{8}*560nm$

= 630 nm

 $\beta =12mm=12*10^{-3}m$

$\begin{array}{l}\beta \text{'}=\frac{\beta }{\mu }=\frac{12*10^{-3}}{4/3}=9*10^{-3}m\\ \beta \text{'}=9mm\end{array}$

$E=\left(I\right)\left(t\right)\left(A\right)?{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta ?$
$\Rightarrow \mathrm{}\left(3.3\right)\left. [\frac{2\pi }{31.4}\right]\left. [3*{10}^{-4}\right]*\frac{1}{2}$

$\Rightarrow \mathrm{}0.99*{10}^{-4}$

Kindly go through the solution 

20

Sol. Angular momentum conservation:

$\begin{array}{rr}& \Rightarrow \mathrm{}{\mathrm{I}}_1{\omega }_1+{\mathrm{I}}_2{\omega }_2=\left({\mathrm{I}}_1+{\mathrm{I}}_2\right){\omega }_{\mathrm{f}}\\ & \Rightarrow \mathrm{}\frac{{\mathrm{M}\mathrm{R}}^2}{2}{\omega }_{\mathrm{o}}=\left(\frac{{\mathrm{M}\mathrm{R}}^2}{2}+\frac{{\mathrm{M}\mathrm{R}}^2}{8}\right){\omega }_{\mathrm{f}}\\ & \Rightarrow \mathrm{}{\omega }_{\mathrm{f}}=\frac{4}{5}{\omega }_{\mathrm{o}}\\ & \Rightarrow \mathrm{}{\mathrm{K}\mathrm{E}}_{\text{final}}=\frac{1}{2}\left({\mathrm{I}}_1+{\mathrm{I}}_2\right){\omega }_{\mathrm{f}}^2=\frac{{\mathrm{M}\mathrm{R}}^2{\omega }^2}{5}\\ & \Rightarrow \mathrm{}\mathrm{K}{E}_{\text{initial}}=\frac{1}{2}\mathrm{I},{\omega }_0^2=\frac{{\mathrm{M}\mathrm{R}}^2{\omega }^2}{4}\\ & \Rightarrow \mathrm{}\mathrm{\%}\text{loss}\Rightarrow 20\mathrm{\%}.\end{array}$

  ${\mu }_2>{\mu }_1$

$\frac{{\mu }_2}{{\mu }_1}>1$               

${\mu }_2=\frac{c}{v_2},{\mu }_1=\frac{c}{v_1}$               

$\frac{{\mu }_2}{{\mu }_1}=\frac{v_1}{v_2}>1\Rightarrow v_1>v_2$               

Frequency remains constant while refraction since energy is constant

$\upsilon =\frac{v}{\lambda }$               

$\lambda \alpha v$               

${\lambda }_1>{\lambda }_2$               

$\lambda \to$ decreases

wavelength and speed decreases but frequency remains constant

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be I= p/4 $\pi r$ ²

 where, r is radius of the wavefront at anytime.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- (a) When a decreases w increases. So, size decreases.

(b) Now, light energy is distributed over a small area and intensity∝1/Area  is decreasing so intensity increases

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation- We know that wavelength of sunlight ranges from 4000 Å to 8000 Å.

Clearly, wavelength λ < width of the slit.

Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed On the screen will be different from the geometrical image.