A screen is placed a distance d = 42.0 cm to the right of a small object. At wha

Question

A screen is placed a distance

d = 42.0 cm

to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +7.00 cmbe placed if the real image of the object is at the location of the screen? (Hint: Use the thin lens equation in the form

s' =

,

so

s'(s f) = sf.

Write

s' = d s

and solve for s in the resulting quadratic equation.)

I know the distances are 8.875 and 33.124 but i do no not what the smallest value of d can be to have the image formed on the screen

Explanation / Answer

distance between screen and object = u +v = 42 cm --------------------1

let the object be placed at u units from lens

so using lens equation

1/f = 1/u + 1/v

f = uv/(u+v)

uv = 7 * 42 = 294 cm

so now use the identity

(u-v)^2 = (u+v)^2 -2uv

u-v)^2 = 42^2 - 4* 294

u -v = 24.24 --------------------------------2

adding 1 and 2

2u = 66.24

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