The left end of a long glass rod 8.00cm in diameter and with an index of refract

Question

The left end of a long glass rod 8.00cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00cm . An object in the form of an arrow 1.40mm tall, at right angles to the axis of the rod, is located on the axis 22.5cmto the left of the vertex of the convex surface.

Part A

Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.

14.77

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Part B

Find the height of the image formed by paraxial rays incident on the convex surface.

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The left end of a long glass rod 8.00cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00cm . An object in the form of an arrow 1.40mm tall, at right angles to the axis of the rod, is located on the axis 22.5cmto the left of the vertex of the convex surface.

Part A

Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.

s? =

14.77

  cm  

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Incorrect; Try Again; 4 attempts remaining

Part B

Find the height of the image formed by paraxial rays incident on the convex surface.

|y?| =   mm  

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Explanation / Answer

apply the formula for focal length of lens as

n1/s1   + n2/s2 = (n2-n1)/R

so here n1 = refractive index of air =1

n2 is refractive index of glass   = 1.6

R is radius = 4 cm

so now by substitutuion, we have

1/22.5   +   1.6/s2 = (1.6-1)/4

1.6/s2   = 0.15 - 0.044

s2 = 1.6/0.1056

s2 = 15.15 cm -----------<<<<<<<<<<<Answer to part A

------------------------------------

use the formula for magnification as m = Image ht/Object ht = image dis/ obj dis

so

m = S'/s

y'/1.4 = 15.15/22.5

y' = 1.4 * 15.15/22.5

y' = 0.942 mm ---------------<<<<<<<Answer to part B

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