Sunlight of intensity 1.3kW\,m^ - 2 is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW\,m^ - 2, at a distance 22 cm from the lens on the other side is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Advanced 2018 · Shift 1 · Q47 with Solution

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Correct answer: 130

Sunlight is effectively a parallel beam, so after passing through the convex lens, rays converge to the focal point.

Let the lens aperture area be $A$ .

Since there is no energy loss, total power crossing the lens is

P = I_0 A.

This same power is distributed over the beam cross-section at $22\,\text{cm}$ .

Parallel rays converge linearly to a point at the focus ( $20$ cm), then diverge linearly beyond it.

If the beam radius at the lens is $R$ , then by similar triangles, at distance $22$ cm from lens, the beam radius $r$ satisfies

\frac{r}{R} = \frac{22-20}{20} = \frac{2}{20} = \frac{1}{10}.

Hence beam area at $22$ cm is

A' = \pi r^2 = \pi \left(\frac{R}{10}\right)^2 = \frac{1}{100}\pi R^2 = \frac{A}{100}.

So the cross-sectional area becomes $1/100$ of the lens aperture area.

Intensity is power per unit area:

I = \frac{P}{A'} = \frac{I_0 A}{A/100} = 100 I_0.

Therefore,

I = 100 \times 1.3 = 130\,\text{kW m}^{-2}.

\boxed{130}

Stored correct answer = $130$

This matches the derived answer.