Single Choice

A thin convex lens made from crown glass $$\displaystyle \left ( \mu = \dfrac{3}{2} \right )$$ has focal length f. When it is measured in two different liquids having refractive indices $$\dfrac{4}{3}$$ and $$\dfrac{5}{3} $$, it has the focal lengths $$f_1$$ and $$f_2$$ respectively. The correct relation between the focal lengths is:

A$$f_2 > f$$ and $$f_1$$ becomes negative
B$$f_1$$ and $$f_2$$ both become negative
C$$f_1 = f_2 < f$$
D$$f_1 > f$$ and $$f_2$$ becomes negative
Correct Answer

Solution

Using lens makers formula and taking ratios,$$\displaystyle \dfrac{f_m}{f} = \dfrac{(\mu - 1)}{\left ( \dfrac{\mu}{\mu_m} - 1\right )}$$

$$\Rightarrow \displaystyle \dfrac{\left ( \dfrac{3}{2} - 1\right )}{\left ( \dfrac{3/2}{4/3} - 1\right )} = 4$$

$$\Rightarrow f_1 = 4f$$

$$\displaystyle \dfrac{f_2}{f} = \dfrac{\left ( \dfrac{3}{2} - 1\right )}{\left ( \dfrac{3/2}{5/3} - 1 \right )} = -5$$

$$\Rightarrow f_2 < 0$$

SIMILAR QUESTIONS

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