A biconvex lens of focal length 15 ,cm is in | vedclass

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

Question

(C) Step $1$: Refraction through the lens.Using the lens formula: $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$.Given $u = -30 \,cm$ and $f = +15 \,cm$.

Vedclass · Accepted Answer

virtual and at a distance of $20 \,cm$ from the mirror

Vedclass · Answer

(C) Step $1$: Refraction through the lens.Using the lens formula: $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$.Given $u = -30 \,cm$ and $f = +15 \,cm$.$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{15} \implies \frac{1}{v_1} = \frac{1}{15} - \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}$.So,$v_1 = +30 \,cm$. This image acts as a virtual object for the mirror.Step $2$: Reflection from the mirror.The distance of this image from the lens is $30 \,cm$ to the right. Since the mirror is $10 \,cm$ from the lens,the image is $30 - 10 = 20 \,cm$ behind the mirror.For a plane mirror,the object distance $u_m = +20 \,cm$ (virtual object).The image formed by the plane mirror will be at $v_m = -20 \,cm$ (real image in front of the mirror).Step $3$: Refraction again through the lens.The light rays reflected from the mirror travel back through the lens. The object for the lens is now at $10 + 20 = 30 \,cm$ to the left of the lens $(u_2 = -30 \,cm)$.Using the lens formula again: $\frac{1}{v_2} - \frac{1}{-30} = \frac{1}{15} \implies v_2 = +30 \,cm$.This image is $30 \,cm$ to the right of the lens,which is $30 - 10 = 20 \,cm$ behind the mirror. Since the light rays are converging towards a point behind the mirror,the final image is virtual and located $20 \,cm$ from the mirror.

$A$ biconvex lens of focal length $15 \,cm$ is in front of a plane mirror. The distance between the lens and the mirror is $10 \,cm$. $A$ small object is kept at a distance of $30 \,cm$ from the lens. The final image is

Similar Questions

$A$ lens is placed on a plane mirror. An object needle and its image coincide at $15 \, cm$. The focal length of the lens is......$cm$.

An equiconvex lens of radius of curvature $R = 20 \, cm$ and refractive index $\mu = 1.5$ is silvered on one side. The equivalent focal length of the system is: (in $, cm$)

$A$ convex lens has power $P$. It is cut into two halves along its principal axis. Further,one piece (out of two halves) is cut into two halves perpendicular to the principal axis as shown in the figure. Choose the incorrect option for the reported lens pieces.

Given is a thin convex lens of glass (refractive index $\mu$) and each side having radius of curvature $R$. One side is polished for complete reflection. At what distance from the lens,an object be placed on the optic axis so that the image gets formed on the object itself?

Vedclass Test Series

Exam Paper Generator

Online Exam Module