Chapter 34 — Geometric Optics

15 points · 3 extra credit

Chapter 34 — Geometric Optics

Core: 15 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

Use the sign conventions of this chapter: for mirrors and lenses, object distance $s$ is positive when the object is on the incoming (light-side) of the surface; image distance $s'$ is positive when the image is on the outgoing side (real image); the radius $R$ is positive when its center of curvature is on the outgoing side. Lateral magnification is $m=-\dfrac{s'}{s}$; $m>0$ means upright, $m<0$ means inverted.

1. 1. (1 pt •) A person stands $1.5\,\text{m}$ in front of a flat bathroom mirror.

   (a) How far behind the mirror is the image, and is it real or virtual?

   (b) What is the lateral magnification, and is the image upright or inverted?

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2. 2. (2 pts ••) A concave makeup mirror has radius of curvature $R=24\,\text{cm}$. A small object $4.0\,\text{cm}$ tall is placed on the axis.

   (a) Find the focal length.

   (b) With the object $36\,\text{cm}$ from the mirror, find the image distance, the magnification, and the image height. State whether the image is real or virtual, upright or inverted.

   (c) Now move the object to $8.0\,\text{cm}$ from the mirror. Recompute the image distance and magnification, and describe the image (this is how the mirror is actually used).

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3. 3. (2 pts ••) A convex security mirror has radius of curvature of magnitude $40\,\text{cm}$. A $6.0\,\text{cm}$-tall object stands $30\,\text{cm}$ in front of it.

   (a) Find the focal length (mind the sign).

   (b) Find the image distance, the magnification, and the image height, and describe the image.

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4. 4. (2 pts ••) A long glass rod ($n=1.50$) ends in a convex spherical surface of radius $R=10\,\text{cm}$ facing the air ($n=1.00$). A small object $3.0\,\text{mm}$ tall sits in the air on the axis, $30\,\text{cm}$ from the vertex of the surface.

   (a) Use the refraction relation for a single spherical surface to find where the image forms inside the glass.

   (b) Find the lateral magnification (use $m=-\dfrac{n_1 s'}{n_2 s}$) and the image height, and state whether the image is real or virtual, upright or inverted.

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5. 5. (2 pts ••) A thin converging lens has focal length $f=15\,\text{cm}$. A $2.0\,\text{cm}$-tall object is placed on its axis.

   (a) With the object $45\,\text{cm}$ from the lens, find the image distance, the magnification, and the image height, and describe the image.

   (b) Now place the same object $10\,\text{cm}$ from the lens. Find the image distance and magnification, and describe the image (this is the lens used as a magnifier).

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6. 6. (2 pts ••) A thin diverging lens has focal length $f=-20\,\text{cm}$. A $5.0\,\text{cm}$-tall object is placed $30\,\text{cm}$ from it.

   (a) Find the image distance and magnification.

   (b) Give the image height and describe the image (real/virtual, upright/inverted, enlarged/reduced).

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7. 7. (2 pts ••) A thin lens is ground from glass of index $n=1.60$.

   (a) The lens is biconvex, with front radius of magnitude $20\,\text{cm}$ and back radius of magnitude $30\,\text{cm}$. Using the sign convention, assign signs to $R_1$ and $R_2$ and use the lensmaker's equation to find its focal length. Is it converging or diverging?

   (b) A second lens is planoconvex, with one flat face and one curved face of radius of magnitude $12\,\text{cm}$, made of the same glass. Find its focal length.

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8. 8. (2 pts ••) Two thin converging lenses are mounted on a common axis, $50\,\text{cm}$ apart. The first has $f_1=15\,\text{cm}$, the second $f_2=12\,\text{cm}$. A $1.0\,\text{cm}$-tall object is placed $30\,\text{cm}$ in front of the first lens. Treating the image formed by the first lens as the object for the second:

   (a) Find the location and magnification of the intermediate image formed by lens 1.

   (b) Find the location of the final image (measured from lens 2).

   (c) Find the overall lateral magnification and the final image height, and state whether the final image is real or virtual, upright or inverted.

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9. Extra Credit (3 pts •••)

   Derive the lensmaker's equation from the refraction relation for a single spherical surface, treating a thin lens of index $n$ (in air) as two refracting surfaces of radii $R_1$ and $R_2$ separated by negligible thickness.

   (a) Apply the single-surface relation to the first surface for an object at distance $s_1$, obtaining the intermediate image distance $s_1'$.

   (b) Argue why, for a thin lens, the object distance for the second surface is $s_2=-s_1'$, then apply the single-surface relation to the second surface (now going from glass back to air) to obtain the final image distance $s_2'$.

   (c) Add the two results to eliminate $s_1'$ and show that

   $$\frac{1}{s_1}+\frac{1}{s_2'}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right),$$

   so that the focal length (the image distance for $s_1\to\infty$) satisfies $\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.

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