NCERT Solutions for Class 12 Physics Chapter 10

The speed of light in glass is not independent of the colour of light. The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

Refractive index of glass, μ = 1.5 Brewster angle = θ Brewster angle is related to refractive index as: tan⁡θ = μ ⇒ θ = tan⁻¹⁡(1.5) = 56.31° Therefore, the Brewster angle for air to glass transition is 56.31°.

No; Wave theory Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged. Hence, we can write the expression: c sin⁡i = v sin⁡r … (i) Where, i = Angle of incidence r = Angle of reflection c = Velocity of light in air v = Velocity of light in water We have the relation for relative refractive index of water with respect to air as: μ = v/c Hence, equation (i) reduces to v/c = sin⁡i/sin⁡r = μ … (2) But, μ>1 Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v. The wave picture of light is consistent with the experimental results.

No Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same. In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

In a single slit diffraction experiment, if the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity of the central diffraction band increases up to four times.

The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.

When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.

Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves. On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.

The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.

Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen.

The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second order wave equation, then any linear combination of y1 and y2 will also be the solution of the wave equation.