3. (a) A spherical mirror is used to form a real image, four times the size of t

Question

3. (a) A spherical mirror is used to form a real image, four times the size of the original object, on a screen that is 3.00 m from the mirror. (i) Draw a rough sketch showing the relative positions of the object, mirror, and screen so that this might be achieved. [41 (ii) How far from the mirror must the object be? [4J (iii) Is the mirror convex or concave? Find its focal length and radius of curvature. [6] (b) An object is placed 15.0cm to the left of a thin converging lens of focal length 10.0 cm. A second thin lens is located 10.0 cm to the right of the first lens, and the final image of the original object is formed halfway between the two lenses (that is, 5.0 cm from each lens). (j) What is the focal length of the second lens? [8] (ii) Specify as completely as you can the size and nature of the final image. [6]

Explanation / Answer

cartesian sign convention is used in the following calculation.
part B:

step 1:
for the first lens:
focal length=f=10 cm
object distance=u=-15 cm
let image distance=v
then using lens equation:
(1/v)-(1/u)=1/f

==> (1/v)=(1/u)+(1/f)

==> v=30 cm

as v is positive, the image is formed to the right of the lens at a distance of 30 cm

as it is a converging lens and the object is beyond the focal distance of the lens, the image is formed at the right side of the lens

and it will be inverted.

as we know, image height/object height=image distance/object distance

==> image height/object height=-2....(1)

step 2:

let the focal length of second lens be f.
for the second lens:
object distance=u=30-10=20 cm

image distance=v=-5 cm
using lens equation:

(1/v)-(1/u)=1/f

we get f=-4 cm

so the second lens is diverging lens and focal length of 4 cm.

image height/object height=image distance/object distance=-0.25

so net magnification achieved=-0.25*-2=0.5

hence the final image is upright and half the height of the original object.

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