Optics Ques 69
- A concave lens of glass, refractive index $1.5$ has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$ , it will behave as a
$(1999,2 M)$
(a) convergent lens of focal length $3.5$ $ R$
(b) convergent lens of focal length $3.0$ $ R$
(c) divergent lens of focal length $3.5$ $ R$
(d) divergent lens of focal length $3.0$ $ R$
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Answer:
Correct Answer: 69.(a)
Solution:
Formula:
Refraction at Spherical Thin Lens:
- $R _1=-R, R _2=+R, \mu _g=1.5$ and $\mu _m=1.75$
$ \therefore \quad \frac{1}{f}=(\frac{\mu _g}{\mu _m}-1 )\quad (\frac{1}{R _1}-\frac{1}{R _2}) $
Substituting the values, we have
$\frac{1}{f} =(\frac{1.5}{1.75}-1) \quad (\frac{1}{-R}-\frac{1}{R})=\frac{1}{3.5 R} $
$\therefore \quad f =+3.5$ $ R$
Therefore, in the medium it will behave like a convergent lens of focal length $3.5$ $R$.
It can be understood as, $\mu _m>\mu _g$, the lens will change its behaviour.
BETA
