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The equiconvex lens of glass of $\mu =1.5$ and focal length $15\;cm$ is silvered on one side. It will behave like.

(a)Concave mirror of focal length 10 cm

(b)Convex mirror of focal length 10 cm

(c)Concave mirror of focal length 3.75 cm

(d)Convxe mirror of focal length 2.5 cm

1 Answer

Since the focal length of equiconvex of $\mu =1.5 \;cm$
$|R_1 |=|R_2|=f=10\;cm$
Power of the combination
$P=2 P_L+P_M$
$-\large\frac{1}{F} =2( \mu-1) \bigg[ \large\frac{1}{R_1} -\frac{1}{R_2}\bigg]-\frac{2}{R_2 }$
$\qquad= 2(1.5 -1) \bigg[ \large\frac{1}{15} -\frac{1}{-15}\bigg] -\frac{2}{-15}$
$\qquad= \large\frac{-1}{F}=\frac{4}{15}$
$F= -\large\frac{-15}{14}$
$\quad= -3.75\;cm$
System behave like concave mirror of focal length $3.75\;cm$
Hence c is the correct answer.
answered Jan 22, 2014 by meena.p
 

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