Question 22: Ray Optics

Question

A plano-convex lens is made of material of refractive index n. When a small object is placed 30 cm 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 10 cm away form the lens. Which of the following statement(s) is(are) true?

Accepted Answer

For refraction through plano - Convex lens,
$$
m=-2=\frac{v}{u}
$$
$$
\begin{array}{ll}
\Rightarrow  v=-2 u \\
	ext { As, }  u=-30 \mathrm{~cm} \\
\Rightarrow  v=60 \mathrm{~cm}
\end{array}
$$
Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$$
\begin{aligned}
\frac{1}{60}-\frac{1}{-30} & =\frac{1}{f} \\
\Rightarrow f  =20 \mathrm{~cm}
\end{aligned}
$$
By lens maker's formula,
$$
\begin{aligned}
\frac{1}{f} & =(n-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight) \\
\Rightarrow \frac{1}{20} & =(n-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\infty}ight)=\frac{n-1}{\mathrm{R}} \\
\Rightarrow n & =1+\left(\frac{\mathrm{R}}{20}ight) ........(i)
\end{aligned}
$$
for poor reflection from convex surface,
$$
\begin{aligned}
& u=-30 \mathrm{~cm}, \\
& v=10 \mathrm{~cm}, f=\mathrm{R} / 2
\end{aligned}
$$
Using mirror formula,
$$
\begin{aligned}
\frac{1}{v}+\frac{1}{u} & =\frac{1}{f}=\frac{2}{\mathrm{R}} ; \\
\Rightarrow \frac{1}{10}+\frac{1}{-30} & =\frac{2}{\mathrm{R}}
\end{aligned}
$$
$$
\begin{array}{llrl}
\Rightarrow  \frac{2}{\mathrm{R}}  =\frac{2}{30} \\
	herefore  \mathrm{R}  =30 \mathrm{~cm}
\end{array}
$$
Substitute $R=30 \mathrm{~cm}$ in equation (i) to get $n=$ 2.5 for $f=20 \mathrm{~cm}$ (inverted image) and $n=1.5$ for $f=60 \mathrm{~cm}$ (erect image).

Answer

The refractive index of the lens is 2.5

Answer

The radius of curvature of the convex surface is 45 cm

Answer

The faint image is erect and real

Answer

The focal length of the lens is 20 cm

Ray Optics

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