An object is placed \(0.1 \mathrm{~m}\) infront of a convex lens of focal length \(20 \mathrm{~cm}\) made of…
An object is placed \(0.1 \mathrm{~m}\) infront of a convex lens of focal length \(20 \mathrm{~cm}\) made of a material of refractive index 1.5. The surface of the lens away from the object is silvered. If the radius of curvature of the silvered surface is \(22 \mathrm{~cm}\), then the distance of the final image from the silvered surface is
\(10 \mathrm{~cm}\)
\(11 \mathrm{~cm}\)
\(12 \mathrm{~cm}\)
\(13 \mathrm{~cm}\)
Solution
Given,
focal length of convex lens, \(f=20 \mathrm{~cm}\)
object distance, \(u=-0.1 \mathrm{~m}=-10 \mathrm{~cm}\)
refractive index of material, \(\mu=1.5\)
radius of curvature of silvered surface, \(R=22 \mathrm{~cm}\)
and focal length of concave mirror,
\(f_m=\frac{-R}{2}=-11 \mathrm{~cm}\)
Therefore, the power of the mirror,
\(P_m=-\frac{1}{f_m} \Rightarrow P_m=\frac{1}{11} \mathrm{D}\) ...(i)
Further, focal length of the lens is \(20 \mathrm{~cm}\).
So, the power of lens, \(P_l=\frac{1}{20} \mathrm{D}\) ...(ii)
When the light after passing through the lens will be reflected back by concave mirror through convex lens again then , image will formed.
Now, the power,
\(P=\) power of mirror + power of lens + power of lens
\(P=P_m+P_l+P_l\)
From Eqs. (i), (ii) and (iii), we get
\(\begin{aligned}
\quad P & =\frac{1}{11}+\frac{1}{20}+\frac{1}{20} \\
\Rightarrow \quad P & =\frac{1}{11}+2\left(\frac{1}{20}\right)=\frac{21}{110}
\end{aligned}\)
The focal length of equivalent mirror,
\(f=-\frac{110}{21} \mathrm{~cm}\)
Now, the object in front of a convex lens,
\(\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{-21}{110}=\frac{1}{v}-\frac{1}{10} \\
& v=-11 \mathrm{~cm}
\end{aligned}\)
Hence, the distance of final image from the silvered surface is \(11 \mathrm{~cm}\).