A point object O is placed on the principle axis of a convex lens of focal length 10 cm at 12 cm from the lens.
A point object O is placed on the principle axis of a convex lens of focal length 10 cm at 12 cm from the lens. When object is displaced 1 mm along the principle axis magnitude of displacement of image is x1. When the lens is displaced by 1 mm perpendicular to the principal axis displacement of image is x2 in magnitude. Find the value of x1/x2 .
2 Answers
U = -12
f = -10
1/v - 1/u = 1/f
1/v + 1/12 = 1/10
1/v = 1/60
\(-\frac{1}{v^2} \)dv or \(\frac{1}{u^2} \)du
du = \(\frac{v^2}{u^2} \)du
\( \frac{v}{u} \)= same path
△u1 = △u2 = 1mm
△u1 = △v2
△v2 = vf - vi
x2 = △v2 + 1mm
x2 = 2mm
\(\frac{x_1}{x_2}=\frac{1}{2} \)
Given u = -10
f = 12
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{v}-(\frac{1}{-10})=\frac{1}{12}\)
\(\frac{1}{v}=\frac{1}{12}-\frac{1}{10}\)
\(\frac{1}{v}=\frac{10-12}{120}\)
\(\frac{1}{v}=\frac{-2}{120}\)
\(\frac{1}{v}=\frac{-1}{60}\)
\(-\frac{1}{v^2}\) dv + \(\frac{1}{u^2}\) du
du = \(\frac{v^2}{u^2}\) du
\(\frac{v}{u}\) = same path
△u1 = △u2 = 1mm
△u1 = △v2
△v2 = vf - vi
x2 = △v2 + 1mm
x2 = 2mm
\(\frac{x_1}{x_2}=\frac{1}{2}\)