Two convex lenses of focal length 20,cm each | vedclass

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

Question

(D) For the first lens $L_1$,the object is at infinity $(u = -\infty)$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v_1}

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) For the first lens $L_1$,the object is at infinity $(u = -\infty)$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v_1} - \frac{1}{-\infty} = \frac{1}{20} \implies v_1 = 20\,cm$.This image $I_1$ acts as a virtual object for the second lens $L_2$.The distance between the lenses is $d = 60\,cm$.The distance of $I_1$ from $L_2$ is $u_2 = +(60 - 20) = +40\,cm$ (since it is behind the lens).For the second lens $L_2$,using the lens formula:$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$$\frac{1}{v_2} - \frac{1}{40} = \frac{1}{20}$$\frac{1}{v_2} = \frac{1}{20} + \frac{1}{40} = \frac{2+1}{40} = \frac{3}{40}$$v_2 = \frac{40}{3} \approx 13.33\,cm$ from $L_2$.The total distance from the first lens $L_1$ is $d + v_2 = 60 + 13.33 = 73.33\,cm$.Wait,re-evaluating the provided diagram: The image $I_1$ is formed at $20\,cm$ from $L_1$. The distance between lenses is $60\,cm$. Thus,$I_1$ is $40\,cm$ in front of $L_2$. So $u_2 = -40\,cm$.$\frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20} \implies \frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \implies v_2 = 40\,cm$.The final image is at a distance of $60 + 40 = 100\,cm$ from the first lens.

Two convex lenses of focal length $20\,cm$ each are placed coaxially with a separation of $60\,cm$ between them. The image of the distant object formed by the combination is at $...........\,cm$ from the first lens.

Similar Questions

Two lenses of power $-15 D$ and $+5 D$ are in contact with each other. The focal length of the combination is

Two convex lenses ($L_1$ and $L_2$) of equal focal length $f$ are placed at a distance $\frac{f}{2}$ apart. An object is placed at a distance $4f$ to the left of $L_1$ as shown in the figure. The final image is at:

The power of an achromatic convergent lens combination consisting of two lenses is $+ 2D$. The power of the convex lens is $+ 5D$. The ratio of the dispersive power of the convex lens to the concave lens will be:

Two thin convex lenses of focal length $f_1$ and $f_2$ are placed in contact with each other. The equivalent power of the lens combination is . . . . . . .

$A$ lens of power $3.5 \, D$ is placed in contact with a lens of power $-2.5 \, D$. The combination behaves as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module