Answer:
[tex]n_l = 1.97[/tex]
Explanation:
given data:
refractive index of lens 1.50
focal length in air is 30 cm
focal length in water is -188 cm
Focal length of lens is given as
[tex]\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
[tex]\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
[tex]\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
focal length of lens in liquid is
[tex]\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
[tex]=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}[/tex]
rearrange fro[tex] n_l[/tex]
[tex]n_l = \frac{n_g f_l}{f_l+f(n_g-1)}[/tex]
[tex]n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}[/tex]
[tex]n_l = 1.97[/tex]
