A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index 5/3 is poured inside the bucket, then the microscope has to be raised by 30 cm to focus the object again. The height of the liquid in the bucket is :

Solution

Physics sir · 4 steps

The 30 cm the microscope moves is the image shift, not the depth. Write the shift and solve for h.

1. Image rises. Light from the object bends away from the normal as it leaves the liquid. So the object looks closer, at the apparent depth h/μ, where h is the real depth of the liquid.
2. Why raise it. The image moves up by the difference between real and apparent depth. The microscope must rise by exactly this shift to stay focused.

   Δ = h - h/μ = h(1 - 1/μ)

3. Put in numbers. With μ = 5/3, the bracket is 1 - 3/5 = 2/5. So 2/5h = 30 cm, which gives h = (30 × 5)/2 = 75 cm.
4. So. The liquid stands 75 cm high, option (c).

The idea to keep. When a slab or liquid is added, the image shifts by h(1 - 1/μ). Any refocusing distance is this shift, never the apparent depth itself.

Answer from the NTA final answer key. Solution by PrepWiser.