Question

The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?

Answer 1

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  $\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

On putting estimate values, we get

⇒  $\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}$

⇒  $\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}$

⇒  $v=1.94 \ cm$

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

$m=\frac{v}{u}$ and $m=\frac{h_{1}}{h_{0}}$

⇒  $\frac{v}{u} =\frac{h_{1}}{h_{0}}$

⇒  $\frac{1.94}{25}=\frac{{h_{1}}}{15}$

⇒  ${h_{1}}=1.16 \ mm$