JEE Main 2025 Apr 8 Shift 2, Physics Q12: Thin Lens and Mirrors

JEE Main2025Apr 8, Shift 2Physics

A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is ______

A
$30$
B
$45$
C
$\frac{60}{7}$
D
$15$

Solution

For thin lenses in contact, $\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ , with the concave lens carrying a negative focal length.

Substituting: $\frac{1}{f} = \frac{1}{30} + \frac{1}{- 20} = \frac{2 - 3}{60} = - \frac{1}{60}$ , so $f = - 60$ cm (net diverging system).

Applying the thin-lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with object distance $u = - 20$ cm: $\frac{1}{v} - \frac{1}{- 20} = \frac{1}{- 60}$ .

$\frac{1}{v} = - \frac{1}{60} - \frac{1}{20} = - \frac{1 + 3}{60} = - \frac{4}{60} = - \frac{1}{15}$ , giving $v = - 15$ cm.

The image is 15 cm from the lens (on the same side as the object, i.e. virtual), matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Physics Q12: Thin Lens and Mirrors

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