Physics, 09.06.2020 11:57 Kitty10101

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 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 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

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Answer from: erica11223344

The position of the 14.77 cm, height of the image is 0.577 mm and image is inverted as magnification is negative.

The formula of image formation by spherical reflecting surface,

$\bold {\dfrac {n_1}{s} = \dfrac {n_2}{s'} = \dfrac {n_2 -n_1}{R}}$

Where,

n1 - index of refraction for air = 1

n2 - index of refraction for glass = 1.60,

s - object distance from vertex to spherical surface = 24 cm = 0.24 m

R - radius = 4 cm = 0.04 m

Put the values in the formula,

$\bold {\dfrac {1}{0.24} = \dfrac {1.6}{s'} = \dfrac {1.6 -1}{0.04}}\\\\\bold { s' = \dfrac {1.6 }{10.83}}\\\\\bold { s' =14.77\ cm}$

The magnification formula,

$\bold {m = -\dfrac {n_1s'}{n_2s} = \dfrac {y'}{y}}$

$\bold {y' = -\dfrac {n_1s'y}{n_2s}}\\\bold {\bold {y' = -\dfrac {n_1s'}{n_2s} = 0.000577\ m = 0.577\ mm }}$

So, the magnification,

$\bold {m = \dfrac {-0.577}{1.5} = -0.38 }$

Therefore, the position of the 14.77 cm, height of the image is 0.577 mm and image is inverted as magnification is negative.

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Answer from: ernestsbaby6181

s' = 14.77 cm

height of image = 0.577 mm

Since magnification is negative, it is inverted

Explanation:

Where

index of refraction (n₁) = 1 for air,

index of refraction for glass (n₂) = 1.60,

object distance from vertex to spherical surface (s) = 24 cm = 0.24 m,

radius (R) is positive since it is in the same direction as the refracted light = 4 cm = 0.04 m

image distance from vertex to spherical surface (s')

Height of image = y'

height of object = y = 1.5 mm = 0.0015 m

Object - image formula for spherical reflecting surface is:

$\frac{n_1}{s}+\frac{n_2}{s'} =\frac{n_2-n_1}{R}\\\frac{1}{0.24}+\frac{1.6}{s'}=\frac{1.6-1}{0.04} \\ \frac{1.6}{s'}=15-4.17=10.83\\ s'=\frac{1.6}{10.83}=0.1477 m=14.77cm$

From the magnification formula:

$m=-\frac{n_1s'}{n_2s}=\frac{y'}{y}\\ y'=-\frac{n_1s'y}{n_2s} =-\frac{1*0.1477*0.0015}{1.6*0.24} =-0.000577m=-0.577mm$

Magnification (m) = $\frac{y'}{y}=\frac{-0.577}{1.5}=-0.38$

Since magnification is negative, it is inverted

Answer from: Quest

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Answer from: Quest

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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground...