HELP WITH PHYSICS!

Question

To work this problem, use the fact that the image formed by the first surface becomes the object for the second surface. The figure below shows a piece of glass with index of refraction n = 1.50

 surrounded by air. The ends are hemispheres with radii R₁ = 2.00 cm

 and R₂ = 4.00 cm,

 and the centers of the hemispherical ends are separated by a distance of d = 8.32 cm.

 A point object is in air, a distance of p = 1.03 cm

 from the left end of the glass.

(a) Locate the image of the object due to refraction at the two spherical surfaces.

_________ cm to the right of the second surface

(b) Is the image real or virtual?

Answer 1

a. With light going through the piece of glass from left to right, the radius of the first surface is positive and that of the second surface is negative according to the sign convention. Thus, R1 = +2.00 cm and R2 = – 4.00 cm. Applying n1/p + n2/q = (n2 − n1)/R to the first surface gives 1.00/1.03 cm + 1.50/q1 = (1.50 − 1.00)/ + 2.00 cm which yields q1 = – 2.07 cm. The first surface forms a virtual image 2.07 cm to the left of that surface and 16.39 cm to the left of the second surface.

The image formed by the first surface is the object for the second surface, so p2 = +16.39 cm.

Applying n1/p + n2/q = (n2 − n1)/R to the second surface gives 1.5/16.39 cm + 1/q2 = (1.00 − 1.50)/ - 4.00 cm

Thus q2 = +29.87 cm to the right of the second surface

b. The final image distance is positive, so the image is real.