Question

Find velocity of spot on screen if a light incident makes an angle 45with plane mirror and omega is 20,rad,/sec and screen is placed at a distance of,10m

Answer 1

it is given that an incident light on plane mirror makes an angle 45° , angular frequency, ω is 20 rad/s and screen is placed at a distance of 10m.

we have to find the velocity of spot on the screen.

let α is angle made by incident ray with normal.

so, x = 10tanα

differentiating both sides with respect to time,

dx/dt = 10sec²α (dα/dt)

we know when change is incident angle is i then change in reflected ray is 2i.

so, dα/dt = 2dθ/dt , where θ is angle made by reflected ray with normal.

so, dx/dt = v = 10sec²α (2 dθ/dt)

= 10sec²α (2ω)

= 10sec²(45°) × 2 × 20 rad/s

= 400 × 2

= 800 m/s

therefore, velocity of spot on screen is 800 m/s.

Answer 2

Answer:

let α is angle made by incident ray with normal.

so, x = 10tanα

differentiating both sides with respect to time,

dx/dt = 10sec²α (dα/dt)

we know when change is incident angle is i then change in reflected ray is 2i.

so, dα/dt = 2dθ/dt , where θ is angle made by reflected ray with normal.

so, dx/dt = v = 10sec²α (2 dθ/dt)

= 10sec²α (2ω)

= 10sec²(45°) × 2 × 20 rad/s

= 400 × 2

= 800 m/s