Physics NCERT Exemplar Solutions Class 12th Chapter Ten

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b, d)

Explanation-Consider the pattern of the intensity shown in the figure

(i) As intensities of all successive minima is zero, hence we can say that two sources S₁ and S₂ are having same intensities.

(ii) As width of the successive maxima (pulses) increases in continuous manner, we can say that the path difference (x) or phase difference varies in continuous manner.

(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very clear pattern on the screen.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation- According to question, there is a hole at point P₂. From Huygen's principle, wave will propagates from the sources S₁ and S₂. Each point on the screen will acts as secondary sources of wavelets. Now, there is a hole at point P₂ (minima). The hole will act as a source of fresh light for the slits S₃ and S₄.

Therefore, there will be a regular two slit pattern on the second screen

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation- For the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength. In a Young's double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filteration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

Explanation- due to refraction from the glass medium there is a phase change of $\pi$

$?t$ = $\frac{o{p}^{\text{'}\text{'}}}{v}$ = $\frac{\frac{d}{cosr}}{\frac{c}{n}}=\frac{nd}{ccosr}$

According to snells law n= sin $i$ /sinr

Cosr = $\sqrt[]{1-{sin}^{2}r}=\sqrt[]{1-\frac{{sin}^2\theta }{n^2}}$

$?t$ = $\frac{nd}{c({\frac{{1-sin}^2\theta }{n^2})}^{1/2}}=\frac{n^{2}d}{c}$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

Phase difference = $?\varnothing$ = $\frac{2\pi }{T}*?t$ = $\frac{2\pi nd}{\lambda }$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

So net difference = $?\varnothing +\pi$ = $\frac{4\pi d}{\lambda }$ ( $1-\frac{1}{n^2}{sin}^2\theta$ )-1/2+ $\pi$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- ( a)

Explanation- given width of slit is 10⁴10^-10m= 10^-6

Wavelength of sunlight varies from 4000A⁰ to 8000A⁰

As the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear i.e., mixing of colours form white patch at the centre.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (c)

Explanation-Consider the diagram the light beam incident from air to the glass slab at Brewster's angle (ip ). The incident ray is unpolarised and is represented by dot (.).

The reflected light is plane polarised represented by arrows.

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- T₂P= T₂O+OP= D+x

T₁P= T₁O-OP= D-x

S₁P $\frac{\mathit{\lambda }}{2}$ = $\sqrt[]{{{(S}_1{T}_1)}^2+{\left(P{T}_1\right)}^2}=\sqrt[]{D^2+{(d-x)}^2}$

S₂P= $\sqrt[]{{{(S}_2{T}_2)}^2+{\left(P{T}_2\right)}^2}=\sqrt[]{D^2+{(d+x)}^2}$

The minima will occur when S₂P- S₁P= (2n-1) $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{D^2+{(d+x)}^2}$ - $\sqrt[]{D^2+{(d-x)}^2}$ = $\frac{\mathit{\lambda }}{2}$

$x=D$

[D²+4D²]^1/2-[D²+0]^1/2= $\frac{\mathit{\lambda }}{2}$

[5D²]^1/2- [D²]^1/2= $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{5}$ D-D= $\frac{\mathit{\lambda }}{2}$

$\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}$ we get D= 0.404 $\mathit{\lambda }$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- resolving power =1/d = $\frac{2sin\beta }{1.22\lambda }$  = d_min= $\frac{1.22\lambda }{2sin\beta }$  where $\lambda$  is the wavelength of light

So d_min= $\frac{1.22*5500*{10}^{-10}}{2sin\beta }$

$\lambda$ = $\frac{12.27}{\sqrt[]{V}}$ = 0.12 $*$ 10^-9m

$\frac{d\text{'}min}{dmin}$ = $\frac{0.12*{10}^{-9}}{5500*{10}^{-10}}$ = 0.2 $*$ 10^-3

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- When angle of incidence is equal to Brewster's angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented By arrows.

Polarisation by reflection occurs when the angle of incidence Is the Brewster's angle

So tani_b = $\frac{{\mu }_1}{{\mu }_2}$  where $\mu$ ₂< $\mu$ ₁

When the light rays travels in such a medium, the critical angle is

Sini_c= $\frac{{\mu }_2}{{\mu }_1}$

Where $\mu$ ₂< $\mu$ ₁

As tani_b > sini_c for large angles i_bc

Thus the polarisation by reflection occurs definitely.