Geometrical OpticsHard
Question
A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :
Options
A.-2
B.+2
C.+2.5
D.-2.5
Solution
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{net}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$${\frac{1}{v} + \frac{1}{30} = (1.5 - 1)\left( \frac{1}{15} - \frac{1}{\infty} \right) + (1.2 - 1)\left( \frac{1}{\infty} + \frac{1}{12} \right) }{\frac{1}{v} + \frac{1}{30} = \frac{1}{30} + \frac{1}{60} }{v = 60 }{m = \frac{v}{u} = \frac{60}{- 30} = - 2}$$
Create a free account to view solution
View Solution FreeMore Geometrical Optics Questions
A converging lens of focal length f1 is placed in front of and coaxial with a convex with a convex mirror of focal lengt...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 ...In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5 cm . Total magnifica...Minimum and maximum distance should be for clear vision of healthy eye...A person can not see the objects clearly placed at a distance more than 40 cm. He is advised to use a lense of power...