An extended object is placed at point O,10 c | vedclass

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

Question

(B) $1$. Focal length of convex lens $(f_1)$:Using the lens maker's formula: $\frac{1}{f_1} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(B) $1$. Focal length of convex lens $(f_1)$:Using the lens maker's formula: $\frac{1}{f_1} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$$\frac{1}{f_1} = (1.5 - 1) \left[ \frac{1}{20} - \left( \frac{1}{-20} ight) ight] = 0.5 	imes \left( \frac{2}{20} ight) = \frac{1}{20}$$\Rightarrow f_1 = +20 \ cm$$2$. Focal length of concave lens $(f_2)$:$\frac{1}{f_2} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$$\frac{1}{f_2} = (1.5 - 1) \left[ -\frac{1}{20} - \frac{1}{20} ight] = 0.5 	imes \left( -\frac{2}{20} ight) = -\frac{1}{20}$$\Rightarrow f_2 = -20 \ cm$$3$. For lens $L_1$:Object distance $u_1 = -10 \ cm$,focal length $f_1 = +20 \ cm$.Using $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \Rightarrow \frac{1}{v_1} - \frac{1}{-10} = \frac{1}{20}$$\frac{1}{v_1} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20} \Rightarrow v_1 = -20 \ cm$Magnification $m_1 = \frac{v_1}{u_1} = \frac{-20}{-10} = 2$$4$. For lens $L_2$:The image formed by $L_1$ acts as an object for $L_2$. The distance between lenses is $10 \ cm$.Object distance $u_2 = -(20 + 10) = -30 \ cm$,focal length $f_2 = -20 \ cm$.Using $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \Rightarrow \frac{1}{v_2} - \frac{1}{-30} = \frac{1}{-20}$$\frac{1}{v_2} = -\frac{1}{20} - \frac{1}{30} = \frac{-3 - 2}{60} = -\frac{5}{60} = -\frac{1}{12} \Rightarrow v_2 = -12 \ cm$Magnification $m_2 = \frac{v_2}{u_2} = \frac{-12}{-30} = \frac{2}{5} = 0.4$$5$. Total magnification:$m = m_1 	imes m_2 = 2 	imes 0.4 = 0.8$

Similar Questions

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

The focal length of a plano-convex lens is equal to its radius of curvature. Find the refractive index of its material.

$A$ parallel beam of white light falls on a convex lens. Images of blue,red,and green light are formed on the other side of the lens at distances $x$,$y$,and $z$ respectively from the pole of the lens. Then:

$A$ bi-convex lens has a radius of curvature for both surfaces equal to $1/6 \ cm$. If this lens is to be replaced by another convex lens having different radii of curvature on both sides $(R_1 \neq R_2)$,without any change in the lens power,then the possible combination of $R_1$ and $R_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module