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Geometrical Optics question

2024 · 27 Jan · Shift 1 · Q84

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Geometrical Optics
2024 · 27 Jan · Shift 1 · Q84

Geometrical Optics question

2024 · 27 Jan · Shift 1 · Q84

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1

Two immiscible liquids of refractive indices

\frac{8}{5}

and

\frac{3}{2}

respectively are put in a beaker as shown in the figure. The height of each column is

6 \mathrm{~cm}

. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is

\frac{\alpha}{4} \mathrm{~cm}

. The value of

\alpha

is

\underline{\hspace{2cm}}

.

JEE Main 2024 (Online) 27th January Morning Shift Physics - Geometrical Optics Question 58 English

Numerical answer

View written solutionFree

Correct answer: 31

Given data

Refractive index of upper liquid: $\mu_1 = \dfrac{8}{5}$
Refractive index of lower liquid: $\mu_2 = \dfrac{3}{2}$
Height of each liquid column: $6\text{ cm}$
Coin is at the bottom of the beaker.

We need the apparent depth of the coin when viewed from air for near-normal incidence.

Principle used

For viewing through plane layers at near normal incidence, the apparent thickness of a layer is:

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

If several layers are stacked, the apparent depth from the top is the sum of apparent thicknesses of each layer.

Apparent thickness of each layer

Upper layer

t_1 = 6\text{ cm}, \quad \mu_1 = \frac{8}{5}

\text{apparent thickness of upper layer} = \frac{6}{8/5} = 6\cdot\frac{5}{8} = \frac{15}{4}\text{ cm}

Lower layer

t_2 = 6\text{ cm}, \quad \mu_2 = \frac{3}{2}

\text{apparent thickness of lower layer} = \frac{6}{3/2} = 6\cdot\frac{2}{3} = 4\text{ cm}

Total apparent depth

d_{\text{app}} = \frac{15}{4} + 4 = \frac{15}{4} + \frac{16}{4} = \frac{31}{4}\text{ cm}

Given that apparent depth is $\dfrac{\alpha}{4}$ cm,

\frac{\alpha}{4} = \frac{31}{4}

So,

\alpha = 31

Final answer

\boxed{31}

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