Monika has to determine the focal9length of a concave mirror and aconv

1.	Monika has to determine the focal9length of a concave mirror and aconvex lens offocal length of about 15 cm each. Sheuses a distant tree as the object andobtains asharp image of the tree, one byone, on a screen. If 11 and 12 are thedistancesbetween the mirror/lens and thescreen in the two cases, thencalculate 11 and 12.Also, state the nature of theirrespective images obtained on thescreen526:46 PM
Answer» 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 : (30 cm, 30 cm) and (inverted, inverted)

Monika has to determine the focal9length of a concave mirror and aconvex lens offocal length of about 15 cm each. Sheuses a distant tree as the object andobtains asharp image of the tree, one byone, on a screen. If 11 and 12 are thedistancesbetween the mirror/lens and thescreen in the two cases, thencalculate 11 and 12.Also, state the nature of theirrespective images obtained on thescreen526:46 PM

Answer»

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 : (30 cm, 30 cm) and (inverted, inverted)