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

Question

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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Diễm Thu 1 year 2021-09-01T19:45:35+00:00 2 Answers 19 views 0

Answers ( )

  1. thnahdat
    0
    2021-09-01T19:46:35+00:00
    Reply

    Answer:

    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:

    [tex]\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[/tex]

    From the magnification formula:

    [tex]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[/tex]

    Magnification (m) = [tex]\frac{y’}{y}=\frac{-0.577}{1.5}=-0.38[/tex]

    Since magnification is negative, it is inverted

  2. Sigrid
    0
    2021-09-01T19:47:18+00:00
    Reply

    Answer:

    a) 14.77 cm

    b) 0.5769 mm

    c) The image is inverted

    Explanation:

    Given:

    Radius, r = 4.00 cm

    Object distance, u = 24.0 cm

    refractive index, n2 = 1.60

    a) To find the position of the image, let’s use the formula:

    [tex] \frac{n_1}{u} + {n_2}{v} = \frac{n_2 – n_1}{r} [/tex]

    [tex] \frac{n_2}{v} = \frac{n_2 – n_1}{r} – \frac{n_1}{u} [/tex]

    [tex] \frac{1.6}{v} = \frac{1.6 – 1}{4.00} – \frac{1}{24} [/tex]

    [tex] \frac{1.6}{v} = 0.15 – 0.0417 [/tex]

    [tex] \frac{1.6}{v} = 0.1083 [/tex]

    [tex] v = \frac{1.6}{0.1083} [/tex]

    v = 14.77 cm

    b) To find the height of the image, let’s use the magnification formula.

    [tex] m = \frac{y_2}{y_1} = \frac{n_1 d_1}{n_1 d_0} [/tex]

    [tex] \frac{y_2}{1.5} = \frac{1* 14.77}{1.60 24} [/tex]

    [tex] \frac{y_2}{1.5} = 0.3846 [/tex]

    [tex] y_2= 0.3846*1.5[/tex]

    y2 = 0.5769 mm

    c) The image is inverted as its height is 0.5769 mm and it is located at a distance of 14.77 cm from the vertex

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