A rectangle ABCD is kept in front of a concave mirror of focal length f with its corners A and B being, respectively, at distances 2f and 3f from the mirror with AB along the principal axis as shown in the figure. It forms an image A'B'C'D' in front of the mirror. What is the ratio of B'C' to A'D'?
Concept:
The mirror equation - If rays emanating from a point actually meet at another point after reflection and/or refraction, that point is called the image of the first point.
Calculation:
ABCD is a rectangle, then AD = BC ---- (1)
Given that corners, A and B are at distances 2f and 3f from the mirror, respectively, along the principal axis, then,
u for corner A = 2f
From mirror equation,
\(\frac 1f = \frac 1v + \frac 1{2f}\)
⇒ v = 2f
Magnification \(m = \frac {h'}h = -\frac vu = -\frac {2f}{2f} = -1\)
This means size of the image is equal to size of the object,
⇒ AD = A'D' ---- (2)
Now, u for corner B = 3f
From the mirror equation,
\(\frac 1f = \frac 1v + \frac 1{3f}\)
⇒ v = \(\frac 32\)f
Magnification, \(m = \frac {h'}h = -\frac vu = -\frac {\frac32f}{3f} = -\frac 12\)
Means size of the image is half to size of the object,
⇒ B'C' = \(\frac 12\)BC ---- (3)
From equations 1, 2 and 3, the ratio of B'C' to A'D' is,
⇒ \(\frac {B'C'}{A'D'} = \frac{\frac12BC}{AD} = \frac12\)
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