Question

A small object is placed in the air, at a distance $45 \,cm$ from a convex refracting surface of radius of curvature $15\, cm$. If the surface separates air-from glass of refractive index $1.5$, then the position of image is

• A. $100\, cm$
• B. $120 \,cm$
• C. $125 \,cm$
• D. $135\, cm$

Answer 1

Answer: (D)

The given situation is shown below

For, refraction through a spherical surface
$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R} $
$\Rightarrow \frac{1.5}{v}-\frac{1}{(-45)}=\frac{1.5-1}{(+15)}$
$\Rightarrow \frac{1.5}{v}+\frac{1}{45}=\frac{0.5}{15} $
$\Rightarrow \frac{1.5}{v}+\frac{1}{45}=\frac{5}{150} $
$\Rightarrow \frac{1.5}{v}+\frac{1}{45}=\frac{1}{30}$
$ \Rightarrow \frac{1.5}{v}=\frac{1}{30}-\frac{1}{45}$
$\Rightarrow \frac{1.5}{v}=\frac{3-2}{90} $
$\Rightarrow \frac{1.5}{v}=\frac{1}{90}$
$\Rightarrow v=1.5 \times 90=+135\, cm$
Image will be formed at right side of separating interface at a distance of $135\, cm$ from it.