A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of π rad.s-2. The reflected ray at the end of (1/4) s must have turned through

Question

A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of π rad.s-2. The reflected ray at the end of (1/4) s must have turned through (A) 90° (B) 45° (C) 22.5° (D) 11.25°

Tardigrade Admin · Accepted Answer

As we know, if a mirror is rotated

Given, angular acceleration,

α = π r a d / s^{2}

Initial angular velocity

= ω_{0} = 0

Time,

t = \frac{1}{4} s

Now, as

θ = ω_{0} t + \frac{1}{2} α t^{2}

= 0 \times \frac{1}{4} + \frac{1}{2} \times π \times (\frac{1}{4})^{2}

= \frac{180}{2 \times 16} = 5.62 5^{\circ}

Since, mirror is turned by

5.62 5^{\circ}

, therefore reflected ray will be turned by

2 \times 5.625 = 11.2 5^{\circ}

Q. A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of $π$ rad. $s^{- 2}$ . The reflected ray at the end of $\frac{1}{4} s$ must have turned through

$9 0^{\circ}$

$4 5^{\circ}$

$22. 5^{\circ}$

$11.2 5^{\circ}$

Q. A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of π rad.s−2. The reflected ray at the end of 41​s must have turned through

90∘

45∘

22.5∘

11.25∘

Q. A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of $π$ rad. $s^{- 2}$ . The reflected ray at the end of $\frac{1}{4} s$ must have turned through

$9 0^{\circ}$

$4 5^{\circ}$

$22. 5^{\circ}$

$11.2 5^{\circ}$