An arrow 2.10 cm long is located 75.0 cm from a lens that has a focal length f =

Question

An arrow 2.10 cm long is located 75.0 cm from a lens that has a focal length f = 31.0 cm

A. If the arrow is perpendicular to the principal axis of the lens, as in the figure (a), what is its lateral magnification, defined as hi/ho?

B. Suppose, instead, that the arrow lies along the principal axis, extending from 73.5 cm to 75.6 cmfrom the lens, as indicated in the figure (b). What is the longitudinal magnification of the arrow, defined as Li/Lo? (Hint: Use the thin-lens equation to locate the image of each end of the arrow.)

Explanation / Answer

I am using coordinate sign convention. Let me know if you want some other sign convention.

a) Object distance u = -75 cm
Focal length f = 31 cm
Let v = image distance
1/v - 1/u = 1/f
Or, 1/v = 1/f + 1/u
Or, 1/v = (f + u)/uf
Or, v = uf/(f + u)------------------(1)
Magnification = v/u = f/(f + u)
= 31/(31 - 75)
= 31/(-44)
= -0.70
Ans: -0.70. Negative sign means that inverted image will be formed.

b) For left end:-
Object distance u1 = -76 cm
Let v1 = image distance
From (1)
v1 = u1f/(f + u1) = -76 * 31/(31 - 76) =52.35cm

For right end:-
Object distance u2 = -74 cm
Let v2 = image distance
From (1)
v2 = u2f/(f + u1) = -74 * 31/(31 - 74) =53.34cm
Length of image = v2 - v = 53.34 - 52.32 = 1.02 cm
Length of object = u2 - u1 = -74 - (-76) = 76 - 74 = 2 cm
Li/L0 = 1.02/2.1 = 0.48 (approx.)

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