The lens equation:

$\overline{)\frac{\mathbf{1}}{\mathbf{f}}{\mathbf{=}}{\mathbf{(}}\frac{{\mathbf{\eta}}_{\mathbf{L}}}{{\mathbf{\eta}}_{\mathbf{m}}}{\mathbf{-}}{\mathbf{1}}{\mathbf{)}}{\mathbf{(}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}{\mathbf{)}}}$, where η_{L} is the refractive index of the lens and η_{m} is the refractive index of the medium.

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |*R*_{1}|=10cm and |*R*_{2}|=15 cm. The lens is made of glass with index of refraction *n*_{glass}=1.5. We will employ the convention that *R*_{1} refers to the radius of curvature of the surface through which light will enter the lens, and *R*_{2} refers to the radius of curvature of the surface from which light will exit the lens.

A) What is the focal length of the lens if it is immersed in water (n_{water} = 1.3)

f= ____________ cm

B) What is the focal length *f* of this lens in air (index of refraction for air is *n*_{air}=1)?

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