JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :

A
$- 2$
B
$+ 2$
C
$+ 2.5$
D
$- 2.5$

Solution

For thin lenses in contact: $\frac{1}{f _{n e t}} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ .

Lens 1 (left planoconvex, $μ = 1.5$ , curved side $R = 15$ cm facing object): $\frac{1}{f _{1}} = (1.5 - 1) (\frac{1}{15} - \frac{1}{\infty}) = \frac{1}{30}$ .

Lens 2 (right planoconvex, $μ = 1.2$ , curved side $R = 12$ cm): $\frac{1}{f _{2}} = (1.2 - 1) (\frac{1}{\infty} + \frac{1}{12}) = \frac{1}{60}$ .

Applying the lens formula with $u = - 30$ cm: $\frac{1}{v} + \frac{1}{30} = \frac{1}{30} + \frac{1}{60} \Rightarrow \frac{1}{v} = \frac{1}{60} \Rightarrow v = 60$ cm.

Magnification $m = \frac{v}{u} = \frac{60}{- 30} = - 2$ .

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