Question

an optical system consists of an optical lens and a concave mirror, the focal length of length is f=20​cm, the distance between objective and lens is 30, the distance between lens and concave mirro is alos 30cm, the radius of concave mirror is 30cm.
​(a) find the location of the final image
​(b)is this a real or virtual image?
​(c)is this an erect or inverted image?
​(d)what is the overall transversal magnification?

Answer 1

Given focal length of lens is 20 cm(positive value means convex lens). Distnace between object and lens is 30 cm

distnace between lens and concave mirror is 30 cm, and radius of concave mirror is 30 cm

focal length of mirror = radius of curvature/2=30/2=15 cm

Let us first determine the the image location formed due to lens:

object distance=u=30 cm

focal length of lens = f=0 cm

Let image distnace measured from lend be v

1/v + 1/u = 1/f

1/v + 1/30=1/20

v = 60 cm

Therefore the image formed from first lens is 60 cm from the lens and is inverted. this image formed will be the object of the concave lens.

Magnification =M= u/v=30/60=0.5

distance of this object from mirror =u'= 60-30=30 cm

focal length of mirror = f'=15cm

Let image distnace measured with respect to mirror be v'

1/f' = 1/u'+1/v'

1/15=1/30+1/v'

v' = 30 cm (exactly at the same location where the image is formed from lens). Also, the image formed is inverted with respect to the image from first lens

Also, magnification =M'= u'/v' = 30/30 =1

a) Final image is formed at 90 cm from the object

b)the image formed formed by lens is real and that formed by mirror is virtual

c)the image formed by first lens is inverted. the image formed by mirror is again inverted. Therefore, the final image is erect

d) Overall magnification = M*M' =0.5*1=0.5