Ray Optics and Optical Instruments - Result Question 45
47. If $f_V$ and $f_R$ are the focal lengths of a convex lens for violet and red light respectively and $F_V$ and $F_R$ are the focal lengths of concave lens for violet and red light respectively, then we have [1996]
(a) $f_V<f_R$ and $F_V>F_R$
(b) $f_V<f_R$ and $F_V<F_R$
(c) $f_V>f_R$ and $F_V>F_R$
(d) $f_V>f_R$ and $F_V<F_R$
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Answer:
Correct Answer: 47. (a)
Solution:
- (a) $\frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})$
According to Cauchy relation
$\mu=A+\frac{B}{\lambda^{2}}+\frac{C}{\lambda^{4}} \ldots$ Hence $f \propto \lambda$.
Hence, red light having maximum wavelength has maximum focal length.
$\therefore f_v<f_r$ and also $F_v>F_r$ as focal length is negative for a concave lens.
