Learn Additional Exercise 10 with Free Lessons & Tips

Wavelength of line emitted by hydrogen,

λ = 6563 Å

= 6563 × 10⁻¹⁰ m.

Star’s red-shift,

Speed of light,

Let the velocity of the star receding away from the Earth be v.

The red shift is related with velocity as:

Therefore, the speed with which the star is receding away from the Earth is 6.87 × 10⁵ m/s.

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:

… (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:

Hence, equation (i) reduces to

But, > 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.

Let an object at O be placed in front of a plane mirror MO’ at a distance r (as shown in the given figure).

A circle is drawn from the centre (O) such that it just touches the plane mirror at point O’. According to Huygens’ Principle, XY is the wavefront of incident light.

If the mirror is absent, then a similar wavefront X’Y’ (as XY) would form behind O’ at distance r (as shown in the given figure).

can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).

(a) Thespeed of light in a vacuum i.e., 3 × 10⁸ m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.

(b) Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.

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.

Wavelength of light used, λ = 6000 nm = 600 × 10⁻⁹ m

Angular width of fringe,

Angular width of a fringe is related to slit spacing (d) as:

Therefore, the spacing between the slits is.

(a) 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.

(b) 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.

(c) 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.

(d) 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.

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

Distance between the towers, d = 40 km

Height of the line joining the hills, d = 50 m.

Thus, the radial spread of the radio waves should not exceed 50 km.

Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:

Z_P = 20 km = 2 × 10⁴ m

Aperture can be taken as:

a = d = 50 m

Fresnel’s distance is given by the relation,

Where,

λ = Wavelength of radio waves

Therefore, the wavelength of the radio waves is 12.5 cm.

Wavelength of light beam, λ = 500 nm = 500 × 10⁻⁹ m

Distance of the screen from the slit, D = 1 m

For first minima, n = 1

Distance between the slits = d

Distance of the first minimum from the centre of the screen can be obtained as:

x = 2.5 mm = 2.5 × 10⁻³ m

It is related to the order of minima as:

Therefore, the width of the slits is 0.2 mm.

(a) 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.

(b) 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.