(3%) Problem 13: An object is placed a distance do in front of a concave mirror
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Question
(3%) Problem 13: An object is placed a distance do in front of a concave mirror with a radius of curvature r= 15 cm. The image formed has a magnification of M-2.1. 50% Part (a) Write an expression for the object's distance, do Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 2 % per attempt) detailed view 123 0 Submit Hint I give up! Hints: 0% deduction per hint. Hi ints remaining: 2 Feedback: deduction per feedback. 50% Part (b) Numerically, what is the distance in cm?Explanation / Answer
a)
1/f = 1/do + 1/di ...(1)
Where f is the focal length, do and di are object distance and image distance respectively.
1/do = 1/f - 1/di
= (di - f) / (f x di)
Taking reciprocal on both the sides,
do = (f x di) / (di - f)
b)
f = r/2 = 15/2
= 7.5 cm
The sign of focal length is positive for a concave mirror.
Magnification = -di/do = 2.1
di = - 2.1 do
From (1),
1/f = 1/do + 1/di
Substituting the values,
1/7.5 = 1/do - 1/(2.1do)
1/7.5 = 1/do x [1 - 1/2.1]
1/7.5 = 0.524 x 1/do
do = 7.5 x 0.524
= 3.93 cm
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