contestada

The cornea behaves as a thin lens of focal length approximately 1.80 cm , although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.00 mm . (Note: The results obtained here are not strictly accurate, because, on one side, the cornea has a fluid with a refractive index different from that of air.)
(Part A) If this focal length is in air, what is the radius of curvature of the back side of the cornea?

Respuesta :

Answer:

[tex]R_2 = 1.86 cm[/tex]

Explanation:

By lens makers formula we know that focal length is given as

[tex]\frac{1}{f} = (\frac{\mu_2}{\mu_1} - 1)(\frac{1}{R_1} - \frac{1}{R_2})[/tex]

now we know that

[tex]\mu_1 = 1[/tex]

[tex]\mu_2 = 1.38[/tex]

[tex]R_1 = 0.5 cm[/tex]

f = 1.80 cm

now from above expression

[tex]\frac{1}{1.80} = (1.38 - 1)(\frac{1}{0.5} - \frac{1}{R_2})[/tex]

[tex]R_2 = 1.86 cm[/tex]

Otras preguntas