What is the magnitude of the focal length f in the

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What is the magnitude of the focal length | f | in the small angle approximation? Assume the index of refraction of the lens is 2.0 and of the medium is 1.0. 1. | f | = 1 4 a 2. | f | = 2 a 3. | f | = 3 4 a 4. | f | = 3 2 a correct 5. | f | = 5 2 a 6. | f | = a 7. | f | = 4 a 8. | f | = 1 3 a 9. | f | = 1 2 a 10. | f | = 3 a Explanation: R 1 is the x -coordinate of the center of the spherical surface A ; i.e. , R 1 = -| R A | = - 3 a . Similarly, R 2 = - a . The lens maker’s formula gives 1 f = ( n - 1) parenleftbigg 1 R 1 - 1 R 2 parenrightbigg = (2 - 1) parenleftbigg 1 - 3 a - 1 - a parenrightbigg = 2 3 a f = 3 2 a . 037(part2of3)10.0points The sign of the focal length is 1. negative 2. positive correct Explanation: From the previous part, we see that f > 0. 038(part3of3)10.0points A B a 3 a O C A C B What will be the magnitude and the sign of the focal length when the lens is flipped over? 1. The magnitude decreases by more than a factor 2, the sign is unchanged. 2. The magnitude increases by less than a factor 2, the sign is opposite. 3. The magnitude increases by less than a factor 2, the sign is unchanged.
mittal (im5936) – reflection refraction imaging HW – yeazell – (58010) 13 4. The magnitude increases by more than a factor 2, the sign is unchanged. 5. The magnitude stays the same, the sign is opposite. 6. The magnitude decreases by less than a factor 2, the sign is unchanged. 7. The magnitude decreases by less than a factor 2, the sign is opposite. of the lens? Assume the object is on the left-hand side of the lens. 8. The magnitude increases by more than a 10 . 3 cm 10 cm What is the magnitude of the focal length of the lens? Assume the object is on the left-hand side of the lens.
factor 2, the sign is opposite. 9. The magnitude stays the same, the sign is unchanged. correct 10. The magnitude decreases by more than a factor 2, the sign is opposite. Explanation: When the lens is flipped over, R 1 = + | R B | , R 2 = + | R A | 1 f = ( n - 1) parenleftbigg 1 R 1 - 1 R 2 parenrightbigg = (2 - 1) parenleftbigg 1 a - 1 3 a parenrightbigg = 2 3 a f = 3 2 a . Digression: Notice that this is a negative sign due to the interchange of | R 1 | and | R 2 | . This compensates for the sign change in R 1 and R 2 , so f stays the same. 039(part1of3)10.0points A lens has an index of refraction of 1 . 34. The lens has a radius of curvature of 10 . 3 cm on its left-hand side and a radius of curvature of 10 cm on its right-hand side. Explanation: Let : n = 1 . 34 , R 1 = 10 . 3 cm , and R 2 = - 10 cm . From the perspective of the object, the ra- dius of curvature of a lens surface is positive if the center of curvature is behind a surface; otherwise it is negative . Applying the lens maker’s equation, 1 f = ( n - 1) parenleftbigg 1 R 1 - 1 R 2 parenrightbigg = ( n - 1) ( R 2 - R 1 ) R 1 R 2 f = R 1 R 2 ( n - 1) ( R 2 - R 1 ) = (10 . 3 cm)( - 10 cm) (1 . 34 - 1) [ - 10 cm - (10 . 3 cm)] = 14 . 9232 cm , with magnitude 14 . 9232 cm . 040(part2of3)10.0points What is the sign of the focal length? 1. negative 2. Cannot be determined. 3. positive correct Explanation:
mittal (im5936) – reflection refraction imaging HW – yeazell – (58010) 14 Since the lens is thicker on the axis than near its outer edge, the focal length f is posi- tive and the lens is convergent.

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