The dotted part of the lens is cut and kept o | vedclass

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

Question

(D) Let the focal length of the original lens be $F$. When parallel rays fall on the system,the image is formed at the focal plane. The coordinate of

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(D) Let the focal length of the original lens be $F$. When parallel rays fall on the system,the image is formed at the focal plane. The coordinate of the image is $(30, -1)$,which means the focal length $f = 30 \, cm$.The system consists of two lens parts. The upper part (cut part) has its optical axis at $y = x = 2.5 \, cm$. The lower part has its optical axis at $y = -y = -y \, cm$.For the upper lens,the incident ray is at $y = 0$. The height of the ray from the optical axis is $h_1 = 0 - 2.5 = -2.5 \, cm$. The image height $y_1$ is given by $y_1 = h_1 \cdot (f/f) = -2.5 \, cm$.For the lower lens,the incident ray is at $y = 0$. The height of the ray from the optical axis is $h_2 = 0 - (-y) = y \, cm$. The image height $y_2$ is given by $y_2 = h_2 \cdot (f/f) = y \, cm$.Since the image is formed at $y = -1 \, cm$,and the rays are focused by both parts,we consider the displacement of the optical axis. The shift of the image from the optical axis is proportional to the distance of the incident ray from the optical axis. Given the symmetry and the final coordinate $(30, -1)$,we find that $y = 4.5 \, cm$ satisfies the geometric displacement.

Similar Questions

The radius of curvature for a convex lens is $40 \ cm$ for each surface. Its refractive index is $1.5$. The focal length will be ...... $cm$.

$A$ convex lens of refractive index $\frac{3}{2}$ has a power of $2.5 \ D$. If it is placed in a liquid of refractive index $2$, the new power of the lens is: (in $D$)

$A$ divergent lens will produce

In the displacement method of a lens,we obtain two images for different positions of the lens. One image is magnified,and the other is diminished. If $m$ is the magnification of the first image,then the focal length of the lens is given by:

The focal length of a lens with a refractive index of $1.5$ in air is $0.30 \ m$. When it is immersed in a liquid of refractive index $\frac{4}{3}$,its focal length in the liquid will be ...... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module