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Physics Question on Spherical Mirrors

If the refractive index from air to glass is 32\frac{3}{2} and that from air to water is 43\frac{4}{3}, then the ratio of focal lengths of a glass lens in water and in air is

A

1:21: 2

B

2:12: 1

C

1:41: 4

D

4:14: 1

Answer

4:14: 1

Explanation

Solution

nw=43   ng=32n _{ w }=\frac{4}{3} \,\,\, n _{ g }=\frac{3}{2}
fa(ng−1)=fw(ngnw−1)f _{ a }\left( n _{ g }-1\right)= f _{ w }\left(\frac{ n _{ g }}{ n _{ w }}-1\right)
fwfa=ng−1ngnw−1=32−13243−1\frac{ f _{ w }}{ f _{ a }}=\frac{ n _{ g }-1}{\frac{ n _{ g }}{ n _{ w }}-1}=\frac{\frac{3}{2}-1}{\frac{\frac{3}{2}}{\frac{4}{3}}-1}
=3−2298−1=129−88=1218=82=\frac{\frac{3-2}{2}}{\frac{9}{8}-1}=\frac{\frac{1}{2}}{\frac{9-8}{8}}=\frac{\frac{1}{2}}{\frac{1}{8}}=\frac{8}{2}
∴fwfa=41\therefore \frac{ f _{ w }}{ f _{ a }}=\frac{4}{1}