Question

A plano-convex lens is made of a 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 $45$ cm
• C. The faint image is erect and real
• D. The focal length of the lens is $20$ cm

Answer 1

Answer: (A), (D)

The focal length of the lens is $20$ cm

Case I,
When refraction obtains,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, $u=-30cm$
so, $\frac{1}{v}-\frac{1}{-30}=\frac{1}{f}$
so, $\frac{1}{v}+\frac{1}{30}=\frac{1}{f}$ ...(1)
as given, image size is double.
$\frac{v}{u}=-2v=-2\times u\Rightarrow v=60cm$
on putting at equation (1),
$\frac{1}{60}+\frac{1}{30}=\frac{1}{f}f=30cm$

As given in a condition of reflection faint image generated.
So,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
here, $v=10cm,u=-30cmandf=R/2$
$\frac{1}{10}+\frac{1}{-30}=\frac{2}{R}$
$R=30cm$

now for focal lenght, $\frac{1}{f}=\left(\frac{{\mu }_L}{{\mu }_{\delta }-1}\right)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{20}=(\frac{n}{1}-1)(\frac{1}{30}-\frac{1}{0})$
$n-1=\frac{3}{2}n=\frac{5}{2}=2.5$
In this observation, faint image is erect and vertual.
Hence, option (A) and (D) is correct.