Asmall lightbulb is 1.06 m from a screen. a) if you have a convex lens with 20 cm focal length, where are the two lens locations that will project an image of the lightbulb onto the screen?
b) what’s the magnification in the first case (smaller distance from the lightbulb)?
c) what’s the magnification in the second case (larger distance from the lightbulb)?

A)Explanation:

Let object distance be u.

Image distance v = 1.06 - u

Focal length = 20 cm = .2 m

Applying lens formula

1 / v - 1 / u = 1 / f

1 /( 1.06-u)  - 1 / u = 1 / .2

5 u² -3.3u - 1.06 =0

It will have two roots

u₁ = .1636 m, u₂ = .8964 m

B ) Magnification

= u = .1636m , v = .8964m

m = .8964 / .1636 = 5.48

C ) When u = .8964m ,v = .1634m

m = .1634 / .8964

=.182

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close