Homework #1 solution, Assignments of Engineering

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dz/dr=-0.01r for r=1, dz/dr=-0. This is a slope and not an angle. If we take the arctangent of the slope, we will get T 1. T 1 =atan(-0.05)=-0.04996 radians (as drawn, T 1 is negative) The ray is in glass, so n 1 =1. 1.52sin(-0.04996)=1.0sinT 2 T 2 =-0.07598 radians (as drawn, T 2 is negative) The angle with respect to the optical axis is T 2 minus the angle of the surface normal which is -0. Now we want to find the 'r for a given 'z. 'z is the distance from the origin to the detector minus the sag 'z=192.295-(-0.125)=192. 'r='z·tan(-0.02602)=-5. They ray started at a height of 5, so the height it hits the detector at is -0.008.

c) Repeat part b, except start with r= z=-0.005r 2 =-0. Find the slope at the ray intercept: dz/dr=-0.01r for r=1, dz/dr=-0. This is a slope and not an angle. If we take the arctangent of the slope, we will get T 1. T 1 =atan(-0.05)=-0.09967 radians (as drawn, T 1 is negative) The ray is in glass, so n 1 =1. 1.52sin(-0.09967)=1.0sinT 2 T 2 =-0.15183 radians (as drawn, T 2 is negative) The angle with respect to the optical axis is T 2 minus the angle of the surface normal which is -0. Now we want to find the 'r for a given 'z. 'z is the distance from the origin to the detector minus the sag 'z=192.295-(-0.5)=192. 'r='z·tan(-0.05216)=-10. They ray started at a height of 10, so the height it hits the detector at is -0.065.

d)

The first surface the ray hits is the curved surface. Repeat the approach from part a) except the sign of the equation is different. z=0.005r 2 =0. Find the slope at the ray intercept: dz/dr=0.01r for r=1, dz/dr=0.

Z

T 1

T 2

This is a slope and not an angle. If we take the arctangent of the slope, we will get T 1. T 1 =atan(0.01)=0.01 radians (as drawn, T 1 is positive) The ray is in air, so n 1 =1. 1.0sin(0.01)=1.52sinT 2 T 2 =0.006579 radians (as drawn, T 2 is positive) T 2 is the angle the ray makes with respect to the surface normal. We need the angle it makes with respect to the optical axis. If we subtract the angle of the surface normal (0.01) from the angle between the ray and the surface normal, we’ll get -0.003421 radians which is the angle the ray makes with the optical axis. Now we want to find the height of the ray when it reaches the back surface of the lens. The back surface is at z=5. The change in z from the ray intercept to the back surface is 5-sag=4.995.

T is negative 'z is positive as drawn, 'r is negative

'r=4.995·tan(-0.003421)=-0. This means that the ray has an r height that is 1-0.01709=0.98291.

At this point, we know that the ray intercept is (z,r)=(5,0.98291), the ray angle is -0.003421, and it is in an index of 1.52. The surface normal of the back surface is parallel to the optical axis, so the angle the ray makes with respect to the surface normal is -0.003421. Using the law of refraction, we find the refracted angle in air. 1.52sin(-0.003421)=1.0sin(Tair ) Tair =-0. Now we want to know the 'z for a 'r=-0.98291 and an angle of -0. 'z=-0.98291/tan(-0.005200)=189. This is the distance from the back surface which is 5 mm from the origin. Therefore, the ray crosses the optical axis at 194.022.

e) The first steps are identical to part d) except we use an r value of 5. z=0.005r 2 =0. Find the slope at the ray intercept: dz/dr=0.01r for r=1, dz/dr=0. This is a slope and not an angle. If we take the arctangent of the slope, we will get T 1. T 1 =atan(0.05)=0.04996 radians (as drawn, T 1 is positive) The ray is in air, so n 1 =1. 1.0sin(0.04996)=1.52sinT 2 T 2 =0.03286 radians (as drawn, T 2 is positive) T 2 is the angle the ray makes with respect to the surface normal. We need the angle it makes with respect to the optical axis. If we subtract the angle of the surface normal (0.04996) from the angle between the ray and the surface normal, we’ll get -0.01710 radians which is the angle the ray makes with the optical axis.

'z

'r

T

At this point, we know that the ray intercept is (z,r)=(5,9.846), the ray angle is -0.03416, and it is in an index of 1.52. The surface normal of the back surface is parallel to the optical axis, so the angle the ray makes with respect to the surface normal is -0.03416. Using the law of refraction, we find the refracted angle in air. 1.52sin(-0.03416)=1.0sin(Tair ) Tair =-0. Now we want to know the 'r for a 'z=189.022 (the distance from the back of the lens to the detector) and an angle of -0.03416. 'r=189.022·tan(-0.03416)=-9. Since the ray at the back of the surface had a height of 9.846 and 'r=-9.826, the ray height on the detector is 0.020.