Question

2. (20 points) A 3.75 mm tall stick bug is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of the lens has no curvature, but the right surface of the lens has radius of curvature of magnitude 13.0 em. The index of refraction of the lens material is 1.70. What is the location and height of the stick bug image formed by this lens? Is it real or virtual? Erect or inverted?

Answer 1

To find the location and height of the image formed by the planoconvex lens, we first calculate its focal length using the lens-maker's equation. Then, plug the calculated value into the thin lens formula and magnification formula, from which we can determine the location and height of the image and whether it's real/virtual and erect/inverted.

Step-by-step explanation:

To answer this question, we first need to determine the focal length (f) of the planoconvex lens, which can be done using the lens maker's equation for a planar and convex surface: 1/f = (n-1)[ 1/R1 - 1/R2]. Here, n = index of refraction for the lens = 1.70, R1 = radius of curvature for the left surface = ∞ due to no curvature, R2 = -radius of curvature for the right surface = -13.0 cm (Negative because it is convex to incident ray).

After we find the focal length, a thin lens formula can be used to find the location of the image: 1/f = 1/v - 1/u. We know the object distance (u) = -22.5 cm (negative since it is on the left side of the lens facing towards the lens). The sign of image distance (v) will tell us whether the image is on the same side of the object (real) or the opposite side (virtual).

Then we can find out the height of the image using the magnification equation, magnification(m)= -v/u = h'/h, where h denotes the height of the object and h' that of the image. By the signs of the image distance and height, we can determine whether the image is erect or inverted.

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