A bi-convex lens is formed with two thin plan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Using the lens maker's formula $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$:For the first plano-convex lens: $\frac{1}{f_1} = (1.5 - 1)(\f

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Using the lens maker's formula $\frac{1}{f} = (n-1)(\frac{1}{R_1} - \frac{1}{R_2})$:For the first plano-convex lens: $\frac{1}{f_1} = (1.5 - 1)(\frac{1}{14} - \frac{1}{\infty}) = \frac{0.5}{14} = \frac{1}{28} \, cm^{-1}$.For the second plano-convex lens: $\frac{1}{f_2} = (1.2 - 1)(\frac{1}{\infty} - \frac{1}{-14}) = \frac{0.2}{14} = \frac{1}{70} \, cm^{-1}$.The equivalent focal length $f_{eq}$ is given by $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{28} + \frac{1}{70} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20} \, cm^{-1}$.Thus,$f_{eq} = 20 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = -40 \, cm$ and $f = 20 \, cm$:$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$.Therefore,$v = 40 \, cm$.

Similar Questions

Find the linear magnification in the given diagram.

$A$ diverging lens with a focal length magnitude of $25\ cm$ is placed at a distance of $15\ cm$ from a converging lens with a focal length magnitude of $20\ cm$. $A$ beam of parallel light falls on the diverging lens. The final image formed is:

The dispersive powers of the materials of two lenses forming an achromatic combination are in the ratio of $4:3$. If the effective focal length of the combination is $+60 \ cm$, then the focal lengths of the lenses should be:

$A$ convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is $2/3$. Their equivalent focal length is $30 \, cm$. Their individual focal lengths in $cm$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module