A screen is placed a distance d = 35.5 cm to the right of a small object. At wha

Question

A screen is placed a distance d = 35.5 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 +7.65 cm be 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' = (sf) / (s-f) , so s'(s f) = sf. Write s' = d s and solve for s in the resulting quadratic equation.)

A) Shorter distance = ? cm

B) Longer distance = ? cm

C) For this lens, what is the smallest value d can have and an image be formed on the screen?

Explanation / Answer

object distance is = s

image distance is = s'

and

focal length f=7.65 cm

and

s+s'=d=35.5 cm

1/s+1/s'=1/f

s'=s*f/(s-f)

(d-s)=s*f/(s-f)

====>

s^2-(s*d)+ (f*d)=0

s^2-35.5*s+(7.65*35.5)=0

s^2 - (35.5)*s +(272.575)=0

the above equation is quadratic eqn

and

solutions are s=24.3 cm or s=11.2 cm

A)

shorter distance =11.2 cm

B)

longer distance =24.3 cm

C)

image distance s'=d-s

=35.5- 11.2 =24.3 cm

and

1/s+1/s'=1/f

object distance s=s'*f/(s'-f)

s=24.3*7.65/(24.3-7.65)

object distance s=11.2 cm

image distance s'=24.3 cm

hence ,

d=s+s'=35.5 cm

d=35.5 cm

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