Question

NEET Physics question.
please explain​

Answer 1

$\checkmark$ Question

A perfectly reflecting mirror has an area of 1 cm² Light energy is allowed to fall on it 1 h at the rate of 10 W cm². The force that acts on the mirror is

$\checkmark$ Given

A perfectly reflecting mirror has an area of 1 cm² Light energy is allowed to fall on it 1 h at the rate of 10 W cm²

$\checkmark$ To Find

The force that acts on the mirror

$\checkmark$ Solution

We know that,

$\red{\sf \longrightarrow p=\dfrac{h}{\lambda}}$

We also know that,

$\blue{\sf \longrightarrow \lambda=\dfrac{c}{v}}$

Hence,

$\sf \longrightarrow p=\dfrac{h}{(c/v)}$

So,

$\sf \longrightarrow p=\dfrac{hv}{c}$

We know that,

$\green{\sf \longrightarrow E=hv}$

Therefore,

$\red{\sf \longrightarrow p=\dfrac{E}{c}}$

According to the question,

We are asked to find the force that acts on the mirror

Given that,

A perfectly reflecting mirror has an area of 1 cm² Light energy is allowed to fall on it 1 h at the rate of 10 W cm²

Hence,

• E = 10 J

On reflecting mirror,

We know that,

$\pink{\sf \longrightarrow F=\Delta \omega}$

Here,

• F = Force
• Δω = change in momentum

Hence,

Change in momentum per second is given as,

$\blue{\sf \longrightarrow F=2p=\dfrac{2E}{c}}$

We know that,

• c = 3 x 10⁸ m/s

By Substituting the values,

We get,

$\green{\sf \implies F=2p=\dfrac{2 \times 10}{3 \times 10^{8}}}$

On further simplification,

We get,

$\orange{\sf \implies F=6.7 \times 10^{-8} \ N}$

Therefore,

$\pink{\sf \star \ Force \ (F)=6.7 \times 10^{-8} \ N}$

Answer 2

Answer:

\checkmark✓ Question

A perfectly reflecting mirror has an area of 1 cm² Light energy is allowed to fall on it 1 h at the rate of 10 W cm². The force that acts on the mirror is

\checkmark✓ Given

A perfectly reflecting mirror has an area of 1 cm² Light energy is allowed to fall on it 1 h at the rate of 10 W cm²

\checkmark✓ To Find

The force that acts on the mirror

\checkmark✓ Solution

We know that,

\red{\sf \longrightarrow p=\dfrac{h}{\lambda}}⟶p=

λ

h

We also know that,

\blue{\sf \longrightarrow \lambda=\dfrac{c}{v}}⟶λ=

v

c

Hence,

\sf \longrightarrow p=\dfrac{h}{(c/v)}⟶p=

(c/v)

h

So,

\sf \longrightarrow p=\dfrac{hv}{c}⟶p=

c

hv

We know that,

\green{\sf \longrightarrow E=hv}⟶E=hv

Therefore,

\red{\sf \longrightarrow p=\dfrac{E}{c}}⟶p=

c

E

According to the question,

We are asked to find the force that acts on the mirror

Given that,

A perfectly reflecting mirror has an area of 1 cm² Light energy is allowed to fall on it 1 h at the rate of 10 W cm²

Hence,

E = 10 J

On reflecting mirror,

We know that,

\pink{\sf \longrightarrow F=\Delta \omega}⟶F=Δω

Here,

F = Force

Δω = change in momentum

Hence,

Change in momentum per second is given as,

\blue{\sf \longrightarrow F=2p=\dfrac{2E}{c}}⟶F=2p=

c

2E

We know that,

c = 3 x 10⁸ m/s

By Substituting the values,

We get,

\green{\sf \implies F=2p=\dfrac{2 \times 10}{3 \times 10^{8}}}⟹F=2p=

3×10

8

2×10

On further simplification,

We get,

\orange{\sf \implies F=6.7 \times 10^{-8} \ N}⟹F=6.7×10

−8

N

Therefore,

\pink{\sf \star \ Force \ (F)=6.7 \times 10^{-8} \ N}⋆ Force (F)=6.7×10

−8

N