Question

You plan to project an inverted image 30.0 cm to the right of an object. You have a diverging lens with focal length -4.00 cm located 5.10 cm to the right of the object. Once you put a second lens at 18.7 cm to the right of the object, you obtain an image in the proper location.

(a) What is the focal length of the second lens?
___cm

(b) Is this lens converging or diverging?

converging,diverging    

(c) What is the total magnification?


(d) If the object is 12.0 cm high, what is the image height?
___cm

Answer 1

Here Diverging lens will make an image of the object.This image is used by the second lens to produce the fianl image.

First of all we will find out the distance to the image produced by the diverging lens.

using lens equation

F=-4 cm

u=-5.10 cm

or

that is ,diverging lens will produce an image at a distance 18.54cm to the left of the diverging lens.

This image will act as the source to the second lens.

Distance between this image and the second lens u2= 18.54-5.10+18.7=32.14cm=-32.14cm

this minus sign is because the image lies on the left side of the second lens.

Distance between the second lens and the final image v2=30-18.7=11.3cm

this is positive because the image lies on the right side of the second lens.

Using lens equation

or

since f2 is positive the second lens should be converging lens.

b)Second lens is converging lens because the focal length is positive .For a converging lens focal length is positive and for a diverging lens it is negative.

c)Total magnification of the combination is

M=M1*M2 where M1 is the magnification of first lens and M2 is the magnification of the second lens.

Linear Magnification of a thin lens is given by

where d is the distance from lens to the object.For real image M is negative and the image is inverted ,For virtual image M is positive and upright.

Here magnification due to diverging lens

magnification due to second converging lens

total magnification

d)We also have an equation

where hi is the height of the image and ho is the height of the object.

given ho=12cm and M=0.5115