Optics... #1

Classical Mechanics Level 5

A transparent solid sphere of radius 2 cm , and density $\rho$ floats in a transparent liquid of density $2\rho$ kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature 8 cm and is silvered to make it a concave mirror as shown in the above figure. When an object is placed at a distance of 10 cm directly above the centre of the sphere C, its final image coincides with it. Find h (as in figure), the height of the liquid surface in the in the beaker from the apex of the bottom. Consider the paraxial rays only.

A paraxial ray is a ray that makes a small angle to the optical axis of the system, and lies close to the axis throughout the system.

Refractive index of sphere is $\frac{3}{2}$ .

Refractive index of liquid is $\frac{4}{3}$ .

Round that answer to nearest integer .

h is measured in cm

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The answer is 15.

1 solution

Harshvardhan Mehta
Feb 14, 2015

h-10 is not understood.can you please explain?

Adarsh Kumar - 6 years, 3 months ago

The image formed after refraction in the second surface of the sphere must be virtual. The position of this image must be at the center of curvature of the concave mirror. This position is at a distance of (h-10) from the lower surface of the sphere.

Harshvardhan Mehta - 6 years, 3 months ago

It's because the entire length of the beaker can be divided into 3 parts -

1.The half of the sphere = 2 cm in height.

2.The image distance = v,

3.The radius of curvature of the concave mirror = 8cm.

Ayan Jain - 6 years, 3 months ago

Harshvardhan Mehta - 6 years, 3 months ago

Yes, why h-10?

Also, on solving the last equation, we get v= 64/13 cm and not 5 cm.

Ayan Jain - 6 years, 3 months ago

can you please once again check your calculations... i did mine once again but got the same answer.. and yeah also the answer is 15 and not 5 .

Harshvardhan Mehta - 6 years, 3 months ago

Yea, by 5 I meant (15-10) for 'v'. And please can you tell us why you took final image distance to be (h-10)?

Calcs -

4/(3v) - 3/16 = 1/12

4/(3v) = (9+4)/48 ==> v = 64/13

Ayan Jain - 6 years, 3 months ago

@Ayan Jain –

Harshvardhan Mehta - 6 years, 3 months ago

@Harshvardhan Mehta – Yea, I meant that ~ I didn't think that we had to estimate it. I think you should edit the question that estimation has to be done haha. So yes, v after refraction from second surface is 5 cm. And okay I figured why H-10. It's simply because H - 8 -2 cm. Sorry for confusion.

If image after refraction at second surface is formed at 5 cm, it should be a real image right? (Correct me if wrong).

One more solution could be if the image after the last refraction is formed at the pole of the concave mirror. Which means that h = 5 + 2 = 7cm. Correct me if I'm wrong.

Ayan Jain - 6 years, 3 months ago

@Ayan Jain – We need to find v. Then one solution would be v passing through the C of curvature of mirror, or it landing on the pole. Thus H can be equal to V+2.

Ayan Jain - 6 years, 3 months ago

@Ayan Jain – The image formed after refraction in the second surface of the sphere must be virtual. The position of this image must be at the center of curvature of the concave mirror. This position is at a distance of (h-10) from the lower surface of the sphere.

Harshvardhan Mehta - 6 years, 3 months ago

@Harshvardhan Mehta – Isn't the image real? @Md Zuhair

Ankit Kumar Jain - 2 years, 10 months ago

Optics... #1

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The answer is 15.

1 solution

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