One end of a cylindrical glass rod is ground to a hemispherical surface of radius R = 2 cm. An object of height h0 = 1 mm is placed on the axis of the rod, u = 8 cm to the left of the vertex. Find the image distance v and the image height hI, when the rod is in air. The indices of refraction of glass and water are 1.50 and 1.33 respectively.
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Thus the position of the image is v = -18.46 cm
Explanation:
Given data:
- Refractive index of water, μ 1 = 1.33
- Refractive index of the hemisphere, μ 2 = 1.5
- Radius of curvature, R = 2 cm
- object distance, u = − 8 cm
Substituting these values in the following formula, we get
μ 2 /v − μ 1 / u = μ 2 −μ 1 / R
1.5 / v - 1.33 / - 8 = 1.5 - 1.33 / 2
1.5 / v = 0.17 / 2 - 1.33 / 8
1.5 / v = 0.68 - 1.33 / 8
1.5 / v = - 0.65 / 8
V = 1.5 x 8/ - 0.65
V = 12 / - 0.65
V = - 18.46 cm
Thus the position of the image is v = -18.46 cm
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