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Ray Optics and Optical Test - 8

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Ray Optics and Optical Test - 8
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  • Question 1
    1 / -0

    A plano-convex lens acts like a concave mirror of 28 cm focal length when its plane surface is silvered and like a concave mirror of 10 cm focal length when its curved surface is silvered. What is the refractive index of the material of the lens?

    Solution

    A plano-convex lens behaves as a concave mirror if its one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur. Let the focal length of silvered lens is F and focal length of mirror is f(m)
    1/F = 1/f + 1/f +1/f(m)
    Plano-convex lens silvered on plane side has f(m) = infinity
    1/F = 2/f + 1/ infinity
    1/F = 2/f
    f = 28××2 = 56 cm
    Plano-convex lens silvered on convex side has f(m) = R/2
    1/F = 2/f + 2/ R
    1/10 = 2/56 + 2/R
    R = 280/9 cm
    Now
    1/f = (μ–1)/R
    if we put all value 
    μ = 1.55

  • Question 2
    1 / -0

    A pin is placed 10 cm in front of a convex lens of focal length 20 cm and refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature of 22 cm. How far from the lens is the final image formed?

    Solution

    The curved silvered surface will behave as a concave lens of focal length fm = R/2 = −22/2 = −11cm = −0.11m
    And hence
    PM = the power =−1/fm = −1/−0.11 = 1/0.11 D
    Further as focal length of lens is 20 cm i.e. 0.2m its power is
    PL=1/f2 = 1/0.2 D
    Now as in image formation, light after passing through the lens will be reflected back by curved mirror through lens again
    P=PL+PM+PL
    =2PL+PM
    P=2/0.2 + 1/0.11 = 210/11 D
    So, focal length of equivalent mirror.
    F = −1/P = 11/210 m = −110/21 cm
    i.e. silvered lens behaves as a concave mirror of focal length (110/21)cm
    So far object at a distance 10 cm in front of it.
    1/v+1/−10 = 21/110
    i.e. v = −11cm
    The image will be 11 cm in front of silvered lens and will be real.

  • Question 3
    1 / -0

    A telescope, when in normal adjustment, has a magnifying power of 6 and the objective and the eye-piece are 14 cm apart The focal lengths of the eye-piece and the objective respectively are

    Solution

    In normal adjustment, m = fo/fe = 6. Therefore fo = 6 fe
    further, fo+fe = 14cm
    or 7fe = 14cm or fe = 2cm
    hence fo = 12cm

  • Question 4
    1 / -0

    Magnifying power of a compound microscope is high if

    Solution

    Angular magnification or magnifying power of compound microscope is defined as ratio of angle made at eye by image formed at infinity to the angle made by object, if placed at distance of distinct vision from an unaided eye.
    Magnification = LD / fo . fe
    where, L-length of the tube of microscope
    As, m ∝ 1/fo
    and m ∝ 1 / fe
    ∴ both eye piece and objective must be of smaller focal lengths, so, that magnification is higher.

  • Question 5
    1 / -0

    The magnifying power of telescope is high if

    Solution

    magnifying power of telescope is directly proportional to fo/fe.
    Hence fo should be large and fe should be small.

  • Question 6
    1 / -0

    A lens of power + 2.0 D is placed in contact with another lens of power – 1.0 D. The combination will behave like

    Solution

    Combined power, P = P1 + P2 = +2 D + (-1 D) = +1 D
    focal length of combination, f = 1/P = 1/1 = 1m = =100 cm
    Since focal length is positive, it will behave as converging lens.

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