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

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$