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### An extended object is placed at point O ...

An extended object is placed at point O, 10 cm in front of a convex lens $L_1$ and a concave lens $L_2$ is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is

(A) 0.4         (B) 0.8         (C) 1.3         (D) 1.6

Solution

For $L_1$,

$\frac{1}{{{f_1}}} = (1.5 - 1)\left( {\frac{1}{{ + 20}} - \frac{1}{{ - 20}}} \right) = \frac{1}{{20}}$

Also, $\frac{1}{{{v_1}}} - \frac{1}{{ - 10}} = \frac{1}{{{f_1}}} = \frac{1}{{20}}$

$\Rightarrow {v_1} = - 20$

Magnification ${m_1} = \frac{{ - 20}}{{ - 10}} = 2$

For $L_2$,

$\frac{1}{{{f_2}}} = (1.5 - 1)\left( {\frac{1}{{ - 20}} - \frac{1}{{ + 20}}} \right) = - \frac{1}{{20}}$

$\frac{1}{{{v_2}}} - \frac{1}{{ - (20 + 10)}} = \frac{1}{{{f_2}}} = - \frac{1}{{20}}$

$\Rightarrow {v_2} = - 12$

Magnification ${m_2} = \frac{{ - 12}}{{ - 30}} = \frac{2}{5}$

Overall magnification $=m_1 \times m_2 = 2\times \frac {2}{5} = 0.8$

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$