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Geometrical Optics question

2023 · 31 Jan · Shift 2 · Q57

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Geometrical Optics
2023 · 31 Jan · Shift 2 · Q57

Geometrical Optics question

2023 · 31 Jan · Shift 2 · Q57

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1

A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index

\frac{5}{3}

is poured inside the bucket, then the microscope has to be raised by

30 \mathrm{~cm}

to focus the object again. The height of the liquid in the bucket is :

A
$50 \mathrm{~cm}$
B
$18 \mathrm{~cm}$
C
$75 \mathrm{~cm}$
D
$12 \mathrm{~cm}$

View written solutionFree

Correct answer: C

Concept used: Apparent depth through a liquid

When an object is viewed from air through a liquid of refractive index $\mu$ , its apparent depth becomes

\text{apparent depth} = \frac{\text{real depth}}{\mu}

So, if the liquid has actual height $h$ , the object appears raised by

\Delta = h - \frac{h}{\mu} = h\left(1-\frac{1}{\mu}\right)

This upward shift in apparent position means the microscope must be raised by the same amount to refocus.

Given data

Refractive index of liquid: $\mu = \frac{5}{3}$
Microscope is raised by: $\Delta = 30\text{ cm}$

Thus,

30 = h\left(1-\frac{1}{\mu}\right)

Substitute $\mu = \frac{5}{3}$

1 - \frac{1}{\mu} = 1 - \frac{3}{5} = \frac{2}{5}

So,

30 = h\cdot \frac{2}{5}

Solve for $h$

h = 30\times \frac{5}{2} = 75\text{ cm}

Check options

A: $50\text{ cm}$
B: $18\text{ cm}$
C: $75\text{ cm}$
D: $12\text{ cm}$

Hence, the correct option is

\boxed{\text{C: }75\text{ cm}}

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