Physics NCERT Exemplar Solutions Class 12th Chapter Ten

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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

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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.

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Explanation- angular resolution of the eye? = 5.8 x 10^-4

linear distance between two dots is l= 2.54/300=0.84 $*$ 10^-2cm

? = l/z

z=l/? = $\frac{0.84*{10}^{-2}}{5.8*{10}^{-4}}$ = 14.5cm

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Explanation- As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm. The diffraction occur if the wavelength of waves is nearly equal to slit width.

As the wavelength of light waves is 7000 *10^-10 m to 4000 $*$ 10^-10m. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

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Explanation- We know that the sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. Due to the large distance the radius of wavefront can be considered as large (infinity) and hence, wavefront is almost plane.

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Explanation- The point image I₁, due to L₁ is at the focal point. Now, due to the converging lens L₂, let final image formed is I which is point image, hence the wavefront for this image will be of spherical symmetry.

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Explanation- When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of

wavefront is in accordance with Huygen's principle.

So, Huygen's principle is valid for longitudinal sound waves also.

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Explanation-All points with the same optical path length must have the same phase.

So – $\sqrt[]{{\mu }_r{\epsilon }_r}AE$   =BC-–  $\sqrt[]{{\mu }_r{\epsilon }_r}CD$

BC= $\sqrt[]{{\mu }_r{\epsilon }_r}$ (CD-AE)

BC>0, si must be greater than AD

But in other figure

–  $\sqrt[]{{\mu }_r{\epsilon }_r}AE=BC–\mathrm{ }\mathrm{ }\sqrt[]{{\mu }_r{\epsilon }_r}CD$

So BC= –  $\sqrt[]{{\mu }_r{\epsilon }_r}\left(CD-AE\right)$

But clearly here BE is less than zero

To proving snells law we know that

BC=ACsin $\theta$ and CD-AE=ACsin $\theta$

So n= sini/sinr