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…

Question

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

\frac{7}{6}

, the magnification becomes

M_{2}

. The magnitude

|\frac{M_{2}}{M_{1}}|

is

Aicharya · Accepted Answer

For reflection from a concave mirror, 1 v + 1 u = 1 f &#8658; 1 v - 1 15 = &#160; - 1 10 1 v = 1 15 - 1 10 = - 1 30 &#8756; v = - 30 Magnification m 1 = - v u = - 2 Now for refraction from lens, 1 v - 1 u = 1 f &#8658; 1 v = 1 10 - 1 20 = 1 20 &#8756; Magnification m 2 = v u = - 1 &#8756; M 1 = m 1 m 2 = 2 &#160; Now when the set-up is immersed in liquid, no effect for the image formed by mirror. We have &#956; L - 1 1 R 1 - 1 R 2 = 1 10 &#8658; 1 R 1 - 1 R 2 = 1 5 When lens is immersed in liquid, 1 f l e n s = &#956; L &#956; S - 1 &#160; 1 R 1 - 1 R 2 = 2 7 &#215; 1 5 = 2 35 &#8756; 1 v - 1 u = 1 f L i q u i d &#8658; 1 v = 2 35 - 1 20 = 8 - 7 140 = 1 140 &#8756; Magnification = - 140 20 = - 7 &#8756; M 2 = 2 &#215; 7 = 14 &#8756; M 2 M 1 = 7

Consider a concave mirror and a convex lens (refractive index = 1 . 5 ) of focal length 10 cm each,…

Solution