Name: Normal Incidence of Light Ray
Uploaded: 2019-07-19
Duration: 4 min 35 s
Description: When a light ray falls on a glass slab, the way it is refracted depends on the ray that is incident on the glass slab. Watch the video to learn more about it.

Question

A short light pulse of energy

E = 7.5 J

falls in the form of a narrow and almost parallel beam on a mirror plate, whose reflection coefficient is

ρ = 0.60

. The angle of incidence is

30 °

. In terms of the corpuscular theory, find the momentum transferred to the plate. Speed of light

c = 3 \times 10^{8} {ms}^{- 1}

.

Accepted Answer

The linear momentum of each photon is given by,

$ρ = \frac{h}{λ}$

We have shown the components of the incident and reflected photon in the following figure,

Let the absorption coefficient be $a$ . As the sum of absorption coefficient is $30 °$ and the reflection coefficient $ρ$ is $1$ , so, we have,

$a = 1 - ρ . . . . . (1)$

The number of photons incident per unit time is,

$N = \frac{Total energy}{Energy of each photon}$

$\Rightarrow N = \frac{E}{h c / λ}$

$\Rightarrow N = \frac{E λ}{h c} . . . . . . . (2)$

The change in the momentum of a photon, when it is reflected is,

$∆ p_{r} = \frac{2 h}{λ} \cos θ . . . . . . (3)$

The number of reflected photons is,

$N_{1} = N ρ$

$\Rightarrow N_{1} = \frac{E λ}{h c} ρ . . . . . . . (4)$

$∴$ The total change in the momentum due to reflection is,

$∆ p_{R} = N_{1} ∆ p_{r}$

$= \frac{2 E}{c} ρcos θ (Upward)$

$∴$ Hence, the momentum transferred to the plate due to reflection is,

$∆ p'_{R} = \frac{2 E}{c} ρcos θ (Downward) . . . . . . . (5)$

Now, the change in the momentum of a photon due to absorption is,

$∆ \vec{p_{a}} = - \frac{h}{λ} (\sin θ \hat{i} + \cos θ \hat{j}) . . . . . . . . (6)$

Number of photons absorbed,

$N_{2} = N a$

$\Rightarrow N_{2} = \frac{E λ}{h c} (1 - ρ) . . . . . . . (7)$

Using $(1)$ and $(2)$ ,

$∴$ The total change in the momentum due to absorption is,

$∆ \vec{p_{A}} = N_{2} ∆ \vec{p_{a}}$

$\Rightarrow ∆ \vec{p_{A}} = - \frac{E}{c} (\sin θ \hat{i} + \cos θ \hat{j}) (1 - ρ)$

$∴$ The momentum imparted to the plate due to absorption is,

$∆ \vec{p'_{A}} = - ∆ \vec{p_{A}}$

$\Rightarrow ∆ \vec{p'_{A}} = \frac{E}{c} (\sin θ \hat{i} + \cos θ \hat{j}) (1 - ρ) . . . . . . . (8)$

$∴$ Hence, the total momentum imparted to the plate is,

$∆ \vec{p} = ∆ p'_{A} + ∆ \vec{p_{R}}'$

$\Rightarrow ∆ \vec{p} = \frac{E}{c} [(1 - ρ) \sin θ \hat{i} - (1 + ρ) \cos θ \hat{j}] . . . (9)$

Students are advised to remember the above result.

$∴$ Its magnitude is given by,

$\Rightarrow |∆ \vec{p}| = \frac{E}{c} \sqrt{{(1 - ρ)}^{2} \sin^{2} θ + {(1 + ρ)}^{2} \cos^{2} θ}$

$\Rightarrow |∆ \vec{p}| = \frac{E}{c} \sqrt{1 + ρ^{2} + 2 ρcos 2 θ}$

$\Rightarrow |∆ \vec{p}| = \frac{7.5}{3 \times 10^{8}} \sqrt{1 + {(0.6)}^{2} + 2 \times 0.6 \times \cos (2 \times 30 °)}$

$\Rightarrow |∆ \vec{p}| = 35 \times 10^{- 9} N s$

$\Rightarrow |∆ \vec{p}| = 35 nN s$

Important Questions on OPTICS

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A short light pulse of energy E=7.5 J falls in the form of a narrow and almost parallel beam on a mirror plate, whose reflection coefficient is ρ=0.60. The angle of incidence is 30°. In terms of the corpuscular theory, find the momentum transferred to the plate. Speed of light c=3×108 ms-1.

Important Questions on OPTICS

Learn and Prepare for any exam you want !