Monika has to determine the focal length of a concave mirror and a convex lens of focal length of about 15 cm each. She uses a distant tree as the object and obtains the sharp image of the tree, one by one on a screen. The distances l1 and l2 between the mirror/lens and the screen in the two cases and the nature of their respective images obtained on the screen are likely to be: (a) (30 cm, 15 cm) and (erect, inverted) (b) (15 cm, 15 cm) and (inverted, inverted) (c) (15 cm, 30 cm) and (inverted, erect) (d) (30 cm, 30 cm) and (inverted, inverted)
Answers
When an object is placed very close to a concave mirror and convex lens, than only an erect virtual image is formed .
But, a virtual image cannot be obtained on screen .
So , image has to be real and inverted .
Applying mirror formula-
1/v + 1/u = 1/f
where, v = image distance
u = object distance
f = focal length
Here, for concave mirror , f = - 15 cm
if u = - 30 cm
1/v - 1/30 = -1/15
=> image distance , v = - 30 cm
=> the distance l1 between mirror and screen = 30 cm
This is possible when object is placed at centre of curvature of mirror , a real and inverted image is formed at centre of curvature .
Applying lens formula-
1/v - 1/u = 1/f
where, v = image distance
u = object distance
f = focal length
Here, for convex lens , f = 15 cm
if u = - 30 cm
1/v + 1/30 = 1/15
=> image distance , v = 30 cm
=> the distance l2 between lens and screen = 30 cm
This is possible when object is placed at 2f of convex lens , a real and inverted image is formed at 2f of lens on other side .
So, correct option is : (d) (30 cm, 30 cm) and (inverted, inverted)