Question

An object of height 3.0 cm is placed vertically on the principal axis of a convex lens. When the object is -37.5 cm, and image of height -2.0 cm is formed at the distance of 25.0 cm from the lens. Next ,the same object is placed vertically at 25.0 cm from the lens. In this situation the image distance v and the height h of the image is(according to the new cartesian sign convention)
a) v = + 37.5 cm; h = +4.5 cm
b) v = - 37.5 cm; h = + 4.5 cm
c) v = + 37.5 cm; h = - 4.5 cm
d) v = - 37.5 cm; h = - 4.5 cm​

Answer 1

Answer:

v=37.5

v=-37.5

h=4.5

h=-4.5

Answer 2

Answer:

c) v=+37.5 cm; h= -4.5 cm

Explanation:

Case 1-

u= -37.5 cm

h' = -2 cm (real and inverted image formed on the other side of lens as the object)

v= +25 cm (image distance is +ve as it is formed in +ve direction, i.e. on other side of lens)

Now by using lens formula,

1/f = 1/v - 1/u

1/f= 1/25 - (-1/37.5) = 1/25+1/37.5

By solving, we get f=+15 cm----(1)

Case 2-

u= -25 cm

f= +15 cm (from (1))

By lens formula, 1/15 = 1/v - (-1/25)

By solving we get v= +75/2

v= +37.5 cm

Now,

h= 3 cm

$h'_{2}$= ?

m= v/u = $h'_{2}$/h

∴ +37.5/ -25= $h'_2$/ +3

By solving we get, $h'_2$= -4.5 cm

[ we can also cross check by seeing that as image distance is positive (+25), image will be formed on other side of lens, real and inverted. And we have got a minus sign in our image height which indicates real and inverted image. So our answer is correct]

Hope it helps...