Consider a concave mirror and a convex lens (refractive index = 1 . 5 ) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1 ), as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image, formed by this combination, has magnification M 1 . When the set-up is kept in a medium of refractive index 7 6 , the magnification becomes M 2 . The magn

Step-by-step solution

For reflection from a concave mirror, 1 v + 1 u = 1 f ⇒ 1 v - 1 15 =   - 1 10 1 v = 1 15 - 1 10 = - 1 30 ∴ v = - 30 Magnification m 1 = - v u = - 2 Now for refraction from lens, 1 v - 1 u = 1 f ⇒ 1 v = 1 10 - 1 20 = 1 20 ∴ Magnification m 2 = v u = - 1 ∴ M 1 = m 1 m 2 = 2   Now when the set-up is immersed in liquid, no effect for the image formed by mirror. We have μ L - 1 1 R 1 - 1 R 2 = 1 10 ⇒ 1 R 1 - 1 R 2 = 1 5 When lens is immersed in liquid, 1 f l e n s = &#9