Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Sol.

$\begin{array}{r}\end{array}$ $\begin{array}{rr}& {\uparrow }_1^{\mathrm{?}}\mathrm{}\text{(1)}\epsilon =3\\ & \text{(2)}\epsilon -\mathrm{I}\mathrm{r}=2.5\mathrm{V}\\ & \Rightarrow \mathrm{}Ir=0.5\\ & \text{Now, IR}=2.5\\ & \Rightarrow \mathrm{}\frac{\mathrm{R}}{\mathrm{r}}=5.\\ & \Rightarrow \mathrm{}\frac{P_R}{R_r}=\frac{I^{2}R}{I^{2}r}=\frac{R}{r}=5\\ & \Rightarrow \mathrm{}{P}_r=\frac{0.5}{5}=0.1\end{array}$

Sol. $U_{\text{initial}}=\frac{k\left(4q\right)\left(q\right)}{(d/2)}+\frac{k\left(q\right)(-q)}{(d/2)}$

$\begin{array}{rr}& \Rightarrow \mathrm{}\frac{6k{q}^2}{d}\\ & \Rightarrow \mathrm{}{U}_{\text{final}}=\frac{4\left(4q\right)\left(q\right)}{\left(\frac{3d}{2}\right)}+\frac{k\left(q\right)(-q)}{(d/2)}\\ & \Rightarrow \mathrm{}\frac{2}{3}\frac{k{q}^2}{d}\\ & \Rightarrow \mathrm{}\mathrm{\Delta }U=\left(\frac{2}{3}-6\right)\frac{k{q}^2}{d}\Rightarrow \frac{-16}{3}\frac{{\mathrm{k}\mathrm{q}}^2}{d}\end{array}$

  ${\mu }_2>{\mu }_1$

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

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

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

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

A single slit diffraction pattern produces a central maximum and diminishing side bands as it interacts with itself and results from the wavefront bending around the edges of the slit. However, the double slit interference pattern forms equally spaced bright and dark fringes, and it is due to the light superposition from two different coherent sources.

The following conditions are needed to observe sustained (stable) interference:

• The two sources should have a constant phase difference, i.e; they should be coherent.
• The light waves need to have almost the same frequencies.
• The sources must emit waves with comparable amplitudes.
• The path difference should be within the coherence length.

Young's Double-Slit Experiment demonstrates the interference phenomenon. It provides strong proof for the wave nature of light. This experiment shows how due to constructive and destructive interference, two coherent light sources create a pattern of bright and dark fringes.

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.