A convex lens and a concave lens, each with a | vedclass

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

Question

(C) For the convex lens: Object distance $u_1 = -8 \,cm$, focal length $f_1 = +4 \,cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) For the convex lens: Object distance $u_1 = -8 \,cm$, focal length $f_1 = +4 \,cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, we get $\frac{1}{v_1} = \frac{1}{4} + \frac{1}{-8} = \frac{1}{8}$, so $v_1 = +8 \,cm$. This image acts as a virtual object for the concave lens.For the concave lens: The distance between the lenses is $d = 6 \,cm$. The image formed by the convex lens is $8 \,cm$ behind it, which is $8 - 6 = 2 \,cm$ behind the concave lens. Thus, for the concave lens, $u_2 = +2 \,cm$ and $f_2 = -4 \,cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$, we get $\frac{1}{v_2} = \frac{1}{-4} + \frac{1}{2} = \frac{1}{4}$, so $v_2 = +4 \,cm$. This means the final image is formed $4 \,cm$ to the right of the concave lens.The object is $8 \,cm$ to the left of the convex lens. The distance between the object and the convex lens is $8 \,cm$, the distance between the lenses is $6 \,cm$, and the distance between the concave lens and the final image is $4 \,cm$. The total distance between the object and the final image is $8 + 6 + 4 = 18 \,cm$.

$A$ convex lens and a concave lens, each with a focal length of $4 \,cm$, are separated by a distance of $6 \,cm$ along their axis. An object is placed $8 \,cm$ before the convex lens. The distance between the object and its final image is (in $\,cm$)

Similar Questions

$A$ glass convex lens of focal length $0.1 \ m$ is cut into two equal parts along its axis. The ratio of the focal length of the new lenses is:

$A$ convex lens of focal length $30\, cm$ and a concave lens of focal length $10\, cm$ are placed coaxially at a distance $d$ apart. Parallel rays are incident on the convex lens. The rays emerging from the concave lens are also parallel. Then $d = $ . . . . . . $cm$.

An achromatic convergent doublet of two lenses in contact has a power of $+2 \ D$. The convex lens has a power of $+5 \ D$. What is the ratio of the dispersive powers of the convergent and divergent lenses?

Two lenses of power $-1.6 D$ and $+2.1 D$ respectively are placed in contact. The focal length of the combination is (in $cm$)

Two identical glass $(\mu_g = 3/2)$ equiconvex lenses of focal length $f$ each are kept in contact. The space between the two lenses is filled with water $(\mu_w = 4/3).$ The focal length of the combination is

Vedclass Test Series

Exam Paper Generator

Online Exam Module