At the interface of two transparent media, light ray experiences both refraction and reflection. Does the angle of reflection de
pend on the angle of refraction? Can you demonstrate the effect of total internal reflection using the rectangular block? Explain your answer and support your answer with a drawing. In which case the shift of the light beam passing through the transparent block equals zero?
1 answer:
Answer:
No, the angle of reflection is independent on the angle of refraction. The total internal reflection is when the angle of incidence is more than the critical angle.
Explanation:
No, the angle of reflection is independent on the angle of refraction. The angle of reflection is only affected by the angle of incidence.
The total internal reflection is when the angle of incidence (I) is more than the critical angle (C). When the light ray pass through the rectangular block, with time, a critical angle will be formed when the refracted ray is exactly 90 degrees to the normal. As soon as the angle of incidence is more than the critical angle, a total internal reflection is formed (as shown in the attached figure).
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The rocket travelled a maximum height at 1.0102 km.
Given,
The acceleration of a rocket (a) = 12 m/s²
The altitude of the rocket (s) = 0.50 km = 0.5×10³m
The maximum height of the rocket (h) = ?
Solution,
A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.
The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²
(a) = Δv/Δt
Where , Δv is change in velocity and Δt is change in time.
The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.
(v)= Δx/Δt
Where,Δx is the change in position and Δt is change in time & v is velocity.
Therefore we know the equation of motion is written as,
v² = u² +2as
Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.
Then putting the value ,
v² = 0 + ( 2× 10 × 0.5×10³)m/s
v² = m/s
v = 100 m/s
Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,
v² = u² - 2(g)h
h = (v²- u² ) / 2g
h = 10,000/2×9.8
h = 510.2 m
So that the rocket travelled the maximum height ,
(h)= (0.5 km + 510.2m)
(h) = 1.0102 km
Hence, the rocket travelled at the maximum height h is 1.0102 km
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