Optics Question 121
Question: A small piece of wire bent into an L shape with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire is
Options:
A) 1 : 2
B) 3 : 1
C) 1 : 3
D) 2 : 1
Show Answer
Answer:
Correct Answer: B
Solution:
Focal length of mirror $ f=\frac{R}{2}=\frac{10}{2}=5cm $
For part PQ : transverse magnification length of image L1 = $ ( \frac{f}{f-u} )\times L _{0} $ = $ ( \frac{-5}{-5-(-20)} )\times L _{0}=\frac{L _{0}}{3} $
For part QR : longitudinal magnification
Length of image $ L _{2}={{( \frac{f}{f-u} )}^{2}}L _{0} $ = $ {{( \frac{-5}{-5-(-20)} )}^{2}}\times L _{0}=\frac{L _{0}}{9} $
Therefore $ \frac{L _{1}}{L _{2}}=\frac{3}{1} $
BETA
