Question

A marine archaeologist looks out the port of a research submarine, as shown. The port is spherically shaped with center of curvature at point C and radius of curvature r. It is made of a material that has an index of refraction of n_w, the same as the index of refraction of seawater, which is greater than n_a, the index of refraction of air. The archaeologist is located to the left of point C and some equipment in the submarine is located behind the archaeologist. The archaeologist can see through the port, but the port also acts as a mirror so the archaeologist can see the reflection of the equipment.

(a) What is the focal length of the mirror?

(b) On the following figure, sketch a ray diagram to locate the position of the image of the equipment formed as a result of the mirror effect.

(c) Based on your ray diagram, check the appropriate spaces below to describe the image of the equipment formed as a result of the mirror effect.

i. Image is: ..........upright inverted

ii. Image is: ............real virtual

iii. Image is: .........larger than the equipment .............smaller than the equipment

The archaeologist also observes a seahorse located outside the port directly in front of the archeologist. Due to refraction of light at the inner surface of the port, the seahorse does not appear to the archaeologist to be at its actual location.

(d) On the figure above, sketch a ray diagram to locate the position of the image of the seahorse formed by the refraction of light at the port.

(e) Based on your ray diagram, check the appropriate spaces below to describe the image of the seahorse, as seen by the archaeologist, formed by the refraction of light at the port.

i. Image is: ...........upright ............inverted

ii. Image is: ............real ..........virtual

iii. Image is: ...........larger than the seahorse ...............smaller than the seahorse

Answer 1

(a) The focal point is half of the radius of curvature … r/2

(b)

(c) Image is Inverted, Real, Smaller.

(d) Very Tricky. This is not a normal lens question. Since the glass of the window has the same index as the water, it is basically just a one sided lens with no front side (front side being the right) and only bends on the way into air but there is no bending when moving from water to glass because the glass–water boundaries are at the same index. Since its not a thin lens, the thin lens rules don’t apply the same. We draw two rays simply based on the laws of refraction.

The ray towards C hits the lens at 0° and would travel straight through. The horizontal ray drawn above refracts at the water–air interface at an unknown angle that we approximate above to get a rough idea. We teachers, as authors of this solution guide, feel this part of the question is a bit nutty.

e. The image formed is upright, virtual and smaller.