Miscellaneous Geometrical Optics Questions

[zorro]

Homework Statement

I have two problems-

1) A plano-covex lens is silvered on its plane-side and then it acts like a concave mirror of focal length 20cm. When the convex side is silvered it acts like a concave mirror of 7 cm focal length. What is the refractive index of the lens?

2) A point source is placed in a tank filled with a liquid of refractive index √2, at its bottom. Now it starts moving in vertically upward direction with a velocity of 1m/s. The rate of change of visible area at the surface, when source is at 2m from the upper surface of liquid will be?

The Attempt at a Solution

1) 1/F = 2/f_l
Given that F = 20cm
so f_l = 40cm

Using the lens makers equation,
We get 1/40 = (μ-1)(1/R)

Now using the second condition,
1/F = -1/f_l¹
here F = 7cm
So f_l¹ = -7cm
R = -14cm

Substituting this in the first equation we get μ = 13/20 which is weird :|

2)

Let r and h be the radius of the circle illuminated and depth of the source resp.
The rate of change of area at any instant dA/dt = 2πrdr/dt
How do I find a relation between r and h when refractive index is given??

Thanks.

 

[rl.bhat]

Using the lens makers equation,
We get 1/40 = (μ-1)(1/R)

So R = 40(μ-1)

In the second case, focal length of the lens is f = R/(μ-1) and focal length of mirror is fm = R/2.
Hence 1/F = 2(μ-1)/R + 2/R = 2μ/R

1/7 = 2μ/40(μ-1)
Mow solve for μ.

In the second problem, when the total internal refraction take place, r/h = tanθ remains constant. Now 1/μ = sinθ . So θ = 45 degrees. So at any time r = h, and r^2 = h^2/(μ^2 - 1). Now take the derivative to find dr/dt.

 

[zorro]

In the second case, focal length of the lens is f = R/(μ-1) and focal length of mirror is fm = R/2.
Hence 1/F = 2(μ-1)/R + 2/R = 2μ/R

There is no convex lens in the second case- the light enters from the plane side, refraction takes place, gets reflected from the silvered convex side, refracts at plane side and emerges out.

So the equation should be 1/F = 1/∞ + 1/f_m + 1/∞
where f_m = R/2

In the second problem, when the total internal refraction take place, r/h = tanθ remains constant. Now 1/μ = sinθ . So θ = 45 degrees. So at any time r = h, and r^2 = h^2/(μ^2 - 1). Now take the derivative to find dr/dt.

I don't understand why will the total internal reflection take place. The question doesnot mention anything about it.

 

[rl.bhat]

There is no convex lens in the second case- the light enters from the plane side, refraction takes place, gets reflected from the silvered convex side, refracts at plane side and emerges out.

So the equation should be 1/F = 1/∞ + 1/f_m + 1/∞
where f_m = R/2

When you silver the convex surface of the Plano convex lens, the system behave like the combination of Plano convex lens + concave mirror + Plano convex lens. The equation you have written represents the combination of rectangular glass slab + convex mirror + rectangular glass slab which is not true.

I don't understand why will the total internal reflection take place. The question doesnot mention anything about it.

The source inside the tank emits the light in all direction. But as you have shown in the diagram, all the light is not coming out of the surface of the tank because of the total internal reflection. The area of the circular patch of light depends on the depth of the source of light from the surface.

 

[zorro]

Thank you very much !