An object is placed 0.40 ,m from one of the t | vedclass

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

Question

(D) The image formed by the first lens acts as an object for the second lens.For the first lens $(L_1)$:Given $u_1 = -0.40 \,m$ and $f_1 = +0.20 \,m$.

Vedclass · Accepted Answer

infinity

Vedclass · Answer

(D) The image formed by the first lens acts as an object for the second lens.For the first lens $(L_1)$:Given $u_1 = -0.40 \,m$ and $f_1 = +0.20 \,m$.Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{0.20} + \frac{1}{-0.40} = 5 - 2.5 = 2.5 \,m^{-1}$.So,$v_1 = \frac{1}{2.5} = 0.40 \,m$.This image is formed $0.40 \,m$ to the right of $L_1$.For the second lens $(L_2)$:The distance between the lenses is $0.30 \,m$. The image from $L_1$ is $0.40 \,m$ to the right of $L_1$,which means it is $0.40 - 0.30 = 0.10 \,m$ to the right of $L_2$.Since the light rays are converging towards a point $0.10 \,m$ behind $L_2$,this acts as a virtual object for $L_2$.Thus,$u_2 = +0.10 \,m$ and $f_2 = -0.10 \,m$ (concave lens).Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-0.10} + \frac{1}{0.10} = -10 + 10 = 0$.Therefore,$v_2 = \infty$.The final image is formed at infinity.

An object is placed $0.40 \,m$ from one of the two lenses $L_1$ and $L_2$ of focal lengths $0.20 \,m$ and $0.10 \,m$ respectively,as depicted in the figure. The separation between the lenses is $0.30 \,m$. The final image formed by this two-lens system is at

Similar Questions

Two thin convex lenses of focal lengths $30 \ cm$ and $10 \ cm$ are placed coaxially, $1 \ cm$ apart. The power of this combination is (in $D$):

Two lenses of power $6 \, D$ and $-2 \, D$ are placed in contact to form a single lens. The focal length of the combination is:

$A$ 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

Concave and convex lenses are placed touching each other. The ratio of the magnitudes of their power is $2:3$. The focal length of the system is $30 \ cm$. Then the focal lengths of the individual lenses are:

$A$ lens maker has to design an achromatic combination of effective power $-0.5 \, D$. The two types of glasses to be used (say $A$ and $B$) have dispersive powers in the ratio $\omega_A : \omega_B = 3 : 5$. Then the lens made from glass $A$ will have a focal length of ....... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module