A screen is placed a distance d = 36.5 cm to the right of a small object. At wha

Question

A screen is placed a distance d = 36.5 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 +6.85 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.)

longer distance ______ cm

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

Explanation / Answer

here,

Distance to object, d = 36.5 cm
focal length, f = 6.85 cm

Fromt hin Lens Equation,
1/f = 1/s + 1/s' but if the separation is d then s' = (d - s )

so 1/f = 1/s + 1/(d -s)
1/f = (d -s +s)/(s*(d - s)) = d/(s*d - s^2)

upon rearranging we get s*d - s^2 = d*f
or d*(s - f) = s^2 or d = s^2/(s - f)

upon diffrentitating, we get (2s*(s - f) - s^2)/(s - f)^2

when this is zero we have minimized value of d so 2s^2 - 2s*f - s^2 = 0
s^2 - 2*f*s = 0
s = 2f
s = 2 * 6.85
s = 13.7 cm ( Smallest)

Now d = s^2/(s - f)
d = 4f^2/(2f -f)
d = 4f
d = 4 * 6.85
d = 27.4 cm ( largest)

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