The left end of a long glass rod 8.60 cm in diameter, with an index of refractio

Question

The left end of a long glass rod 8.60 cm in diameter, with an index of refraction 1.57, is ground and polished to a convex hemispherical surface with a radius of 4.30 cm . An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 23.0 cm to the left of the vertex of the convex surface. Find the position of the image of the arrow formed by paraxial rays incident on the convex surface. Find the height of the image of the arrow formed by paraxial rays incident on the convex surface.

Explanation / Answer

For a spherical refracting surface
n1 / p + n2 / i = ( n2 - n1 ) / r
Where p and i are the object distance and the image distances respectively. n1 is the refractive index of the first medium and n2 is the refractive index of the second medium. r is the radius of the refracting surface
Here n1 (air) = 1 n2(glass) = 1.5
p = 23 cm
r = 4.30 cm (positive for a convex surface)
1 / 23 + 1.5 / i = 0.5 / 4.30
1.5 / i = 0.116 - 0.043
i = 1.5 / 0.073
i = 20.64 cm
The image is real
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The magnification of the image
m = - i / p = h' / h
Where h' is the height of the image h is the height of the object
m = - 20.64 / 23 = - 0.897
The negative sign shows that the image is inverted
h' / h = 0.897
h' = 1.50 x 0.897 = 1.3455 mm
Thus the height of the image is 1.3455 mm and is inverted

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