Please help! Thnx An arrow 2.00 cm long is located 75.5 cm from a lens that has

Question

Please help! Thnx

An arrow 2.00 cm long is located 75.5 cm from a lens that has a focal length f = 30.5 cm .

Part 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?

m=

Part B Suppose, instead, that the arrow lies along the principal axis, extending from 75.0 cm to 77.0 cm from the lens.

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.)

m=

Explanation / Answer

a) Object distance u = -75.5 cm
Focal length f = 30.5cm
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)
= 30.5/(30.5 - 75.5)
=-.678
Ans: -0.678. Negative sign means that inverted image will be formed.

b) For left end:-
Object distance u1 = -77 cm
Let v1 = image distance
From (1)
v1 = u1f/(f + u1) = -77 * 30.5/(30.5 - 77) = 50.50 cm

For right end:-
Object distance u2 = -75 cm
Let v2 = image distance
From (1)
v2 = u2f/(f + u1) = -51.40 cm
Length of image = v2 - v = 51.40-50.50 = .9 cm
Length of object = u2 - u1 = -75 - (-77) = 2 cm
Li/L0 = .9/2 = 0.45

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