- Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses.
- A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is $\frac{3}{4}^{th}$ of the angle of prism. Calculate the speed of light in the prism.
CBSE OUTSIDE DELHI - SET 1 2017
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Two thin lenses, of focal length $f_1$ and $f_2$ are kept in contact. Let O be the position of object and let u be the object distance. The distance of the image (which is at $I_1$), for the first lens is $v_1$. This image serves as object for the second lens.
Let the final image be at I. We then have
$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u}$
$\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v_1}$
Adding, we get
$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\therefore\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
$\therefore P=P_1+P_2$
At minimum deviation
$r=\frac{A}{2}=30^\circ$
We are given that
$i=\frac{3}{4}A=45^\circ$
$\therefore\text{ }\mu=\frac{\sin 45^\circ}{\sin 30^\circ}=\sqrt{2}$
$\therefore$ Speed of light in the prism $=\frac{c}{\sqrt{2}}$
$(\cong2.1\times10^8 \text{ms}^{-1})$
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