A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in…

1. The refractive index of the lens is 2.5
2. The radius of curvature of the convex surface is 45 cm
3. The faint image is erect and real
4. The focal length of the lens is 20 cm.

Solution

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{60}-\frac{1}{\left(-30\right)}=\frac{1}{f}\Rightarrow f=20cm$ ...(i)
Also $\frac{1}{f}=\left(n-1\right) \left(\frac{1}{R}-\frac{1}{\infty }\right)=\frac{\left(n-1\right)}{R}$ ....(ii)
For reflection from convex mirror (curved surface)

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{R}$ ....(iii)
R = 30 cm
From (i), (ii) and (iii)
$n=2.5,$
Faint image erect and virtual

Asked in: JEE Advanced 2016 (Paper 1)

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