The left end of a long glass rod, 6.00 in diameter, has a convex hemispherical s

Question

The left end of a long glass rod, 6.00 in diameter, has a convex hemispherical surface 3.00 in radius. The refractive index of the glass is 1.60.
A. Determine the position of the image of an object placed in air on the axis of the rod infinitely far from the left end of the rod.

Express your answer in centimeters to three significant figures.

B.Determine the position of the image if an object is placed in air on the axis of the rod 12.0 to the left of the end of the rod.

Express your answer in centimeters to three significant figures.
C. Determine the position of the image if an object is placed in air on the axis of the rod 4.00 from the left end of the rod.

Express your answer in centimeters to three significant figures.

Explanation / Answer

With the object at infinity, the image will be located inside the rod at the effective focal distance f from the end. For this situation you can use a simplified formula, f = r*N2/(N2-N1) = 0.13333 m from the front of the lens. Derivation of the formula above relies on small-angle approximations: The incident ray is parallel to the optical axis. Define ?1 as incident ray angle to the normal of the lens's surface, ?2 refracted ray angle to the normal, ?3 as the angle of the refracted ray relative to the optical axis, and focal distance f as the distance from the front of the lens to the point where the refracted ray intersects the optical axis. The equations that yield f for any incident ray are ?2 = arcsin(sin?1*N1/N2) (?1 is the incidence angle and also the central angle of the ray-lens incidence point.) x1 = r*cos?1; y1 = r*sin?1 (location of incidence point) ?3 = ?1-?2 x2 = y1/tan?3 (hor. distance, incidence point to focal point) f = x2-x1+r As the distance from the incident ray to the optical axis approaches 0, ?2 approaches ?1*N1/N2. ?3 = ?1-?2 which approaches ?1(1-N1/N2). Thus ?1/ ?3 approaches 1/(1-N1/N2) = N2/(N2-N1). With these small angles, r?1 approaches f?3. Thus f approaches r?1/?3 = r*N2/(N2-N1). Note that spherical aberration (ref.) tends to blur the image in a thick spherical lens. With increasing distance from the incident ray to the optical axis, the focal point moves closer to the front of the lens. So a higher quality image is obtained if the rays are irised to occupy only a small central part of the lens's area.

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