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Light Reflection and Refraction Test - 31

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Light Reflection and Refraction Test - 31
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  • Question 1
    1 / -0
    The image of an object placed in front of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distant from the mirror than the object. The magnification of the image is:
    Solution
    We have: $$ \dfrac {1}{f} = \dfrac {1}{u} + \dfrac{1}{v}$$
    Given: $$f = 12\ cm$$ and $$ v = u + 10$$
    $$\implies \dfrac {1}{12} = \dfrac {1}{u} + \dfrac {1}{u +10}$$
    $$\implies u = 20\ cm$$
    $$\implies v = 20 +10 = 30\ cm$$
    We have magnification, $$ m = \dfrac {v}{u} = \dfrac{30}{20} = 1.5$$
  • Question 2
    1 / -0
    An object is situated at a distance of 15cm from a convex lens of focal length 30 cm. The position of the image formed by it will be 
    Solution
    $$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$$

    $$\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{30}$$

    $$\dfrac{1}{v}=\dfrac{1}{30}-\dfrac{1}{15}$$

    $$\dfrac{1}{v}=\dfrac{-1}{30}$$

    $$v= -30cm $$
  • Question 3
    1 / -0
    A plane mirror reflects a beam of light to form a real image. The incident beam is :
    Solution

    A convergent light beam can do this.
    So the answer is "convergent".
    You should be able to show this by drawing a "ray diagram" for the light beam. The reflected rays converge and that's what's needed to produce a real image.
     
    A "real object" always produces a "virtual image" in a plane mirror. That's because the reflected rays diverge.
    Interestingly, a plane mirror can form a real image if the object itself is virtual.
    i.e. rays are convergent.
  • Question 4
    1 / -0
    In the case of refraction of light :
    a) Frequency changes
    b) Speed changes
    c) Wavelength changes
    Solution
    In the process of refraction, speed of light and wavelength of light changes. 

    In a medium, speed of light is given by $$v=\dfrac{c}{\mu }$$  where $$\mu $$ is refractive index of the medium. Hence, for medium with different refractive index , speed of light is different. 

    Frequency will not change because at the boundary/interface of the medium, the number of waves you send is the number of waves you receive at the other side, almost instantly. 

    Since, for a travelling wave we have a relation $$ \nu = \dfrac{v}{\lambda} $$. 
    Hence wavelength also changes in the process of refraction.

    So, option B is correct.
  • Question 5
    1 / -0
    The focal length of a spherical mirror is:
    Solution
    Focal length is the property of a spherical mirror. It does not depend on the light used. So, it is same for all lights.
  • Question 6
    1 / -0
    An object is placed at a distance $$2 f$$ from the pole of a convex mirror of focal length $$f$$ . The linear magnification is:
    Solution

    Here,


    Object distance, $$u=-2f$$

    Image distance, $$v=$$?

    From mirror formula, 

    $$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$


    $$\dfrac{1}{v}+\dfrac{1}{-2f}=\dfrac{1}{f}$$


    $$\dfrac{1}{v}=\dfrac{3}{2f}$$


    $$v=\dfrac{2f}{3}$$

    Now,
    Magnification, $$M=\dfrac{-v}{u}$$


    $$=-\dfrac{\dfrac{2f}{3}}{(-2f)}$$


    $$=\dfrac{1}{3}$$


    (M is positive for convex mirror)

  • Question 7
    1 / -0

    The line PO is called the _______.

    Solution
    An incident ray is a ray of light that strikes a surface.
    Hence the line PO is called the incident ray
  • Question 8
    1 / -0
    The line OQ is called the :

    Solution
    The line OQ is called the reflected ray because incident PO ray is reflected by mirror and  goes along OQ.
  • Question 9
    1 / -0
    It is desired to photograph the image of an object placed at a distance of $$3$$ m from a plane mirror. The camera, which is at a distance of $$4.5$$ m from the mirror should be focused for a distance of :
    Solution
    The distance between the object and the mirror is $$ 3 m$$, i.e., $$d_1 = 3 m$$,

     Hence image distance also equals $$d_1 = 3 m$$

    Also, the distance between the mirror and the camera is $$ 4.5 m$$, i.e., $$d_2 = 4.5 m $$ . 

    Hence the focus of the camera,
    $$ f = d_1 + d_2 = 3 + 4.5 = 7.5 m$$

  • Question 10
    1 / -0
    An object $$5$$ cm tall is placed $$1$$ m from a concave spherical mirror which has a radius of curvature of $$20$$ cm. The size of the image is:
    Solution
    Here,

    Object distance, $$u=-1\ m$$

    Focal length, $$f=-10\ cm$$

    Using mirror formula, 

    $$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$$, we get:

    $$\dfrac{1}{-100}+\dfrac{1}{v}=\dfrac{1}{-20}$$

    or, $$\dfrac{1}{v}=\dfrac{-9}{100}$$

    or, $$v=\dfrac{-100}{9}$$

     Magnification, $$M=\dfrac{-v}{u}=\dfrac{h_{2}}{h_{1}}$$

    or, $$-\dfrac{\dfrac{100}{9}}{100}=\dfrac{h_{2}}{h_{1}}$$

    or, $$h_{2}=\dfrac{-5}{9}=-0.55\ cm$$
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