Answer:
a
The focal length of the lens in water is [tex]f_{water} = 262.68 cm[/tex]
b
The focal length of the mirror in water is [tex]f =79.0cm[/tex]
Explanation:
From the question we are told that
The index of refraction of the lens material = [tex]n_2[/tex]
The index of refraction of the medium surrounding the lens = [tex]n_1[/tex]
The lens maker's formula is mathematically represented as
[tex]\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
Where [tex]f[/tex] is the focal length
[tex]n[/tex] is the index of refraction
[tex]R_1 and R_2[/tex] are the radius of curvature of sphere 1 and 2 of the lens
From the question When the lens in air we have
[tex]\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
When immersed in liquid the formula becomes
[tex]\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
The ratio of the focal length of the the two medium is mathematically evaluated as
[tex]\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }[/tex]
From the question
[tex]f_{air }[/tex]= 79.0 cm
[tex]n_2 = 1.55[/tex]
and the refractive index of water(material surrounding the lens) has a constant value of [tex]n_1 = 1.33[/tex]
[tex]\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}[/tex]
[tex]f_{water} = 262.68 cm[/tex]
b
The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.
