Question

A plano-convex lens is made of material of refractive index $n$. When a small object is placed $30cm$ away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of $10cm$ away from the lens. which of the following statement(s) is (are) true?

• A. The refractive index of the lens is $2.5$
• B. The radius of curvature of the convex surface is $45cm$
• C. The faint image is erect and real
• D. The focal length of the lens is $20cm$ A lens is made of flint glass (refractive index $=1.5$ ).

Answer 1

Answer: (A), (D)

Case 1

Using lens formula,
$\frac{1}{60}+\frac{1}{30}=\frac{1}{f_1}$
$\Rightarrow \frac{1}{f_1}=\frac{1}{60}+\frac{2}{60}$
$\Rightarrow f_1=20cm$
Further, $\frac{1}{f_1}=(n-1)\left(\frac{1}{R}-\frac{1}{\infty }\right)$
$\Rightarrow f_1=\frac{R}{n-1}=+20cm$
Case 2
Using mirror formula,
$\frac{1}{10}-\frac{1}{30}=\frac{1}{f_2}$
$\Rightarrow \frac{3}{30}-\frac{1}{30}=\frac{1}{f_2}=\frac{2}{30}$
$f_2=15=\frac{R}{2}$
$\Rightarrow R=30$
$\Rightarrow R=30cm$
$\frac{R}{n-1}=+20cm=\frac{30}{n-1}$
$\Rightarrow 2n-2=3$
$\Rightarrow f_1=+20cm$
Refractive index of lens is $2.5$.
Radius of curvature of convex surface is $30cm$.
Image is erect and virtual. Focal length of lens is $20cm$.