Question

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

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

Answer 1

Answer: (D)

The given situation is shown below

For, refraction through a spherical surface
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\Rightarrow \frac{1.5}{v}-\frac{1}{(-45)}=\frac{1.5-1}{(+15)}$
$\Rightarrow \frac{1.5}{v}+\frac{1}{45}=\frac{0.5}{15}$
$\Rightarrow \frac{1.5}{v}+\frac{1}{45}=\frac{5}{150}$
$\Rightarrow \frac{1.5}{v}+\frac{1}{45}=\frac{1}{30}$
$\Rightarrow \frac{1.5}{v}=\frac{1}{30}-\frac{1}{45}$
$\Rightarrow \frac{1.5}{v}=\frac{3-2}{90}$
$\Rightarrow \frac{1.5}{v}=\frac{1}{90}$
$\Rightarrow v=1.5\times 90=+135cm$
Image will be formed at right side of separating interface at a distance of $135cm$ from it.