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Light Reflection and Refraction Test - 49

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Light Reflection and Refraction Test - 49
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  • Question 1
    1 / -0
    Mark the correct statement
    Solution
    Pole is the center of a spherical mirror which lies on the principal axis, and is represented by P. Hence none of the options is correct.
  • Question 2
    1 / -0
    Transparent medium bounded by two refracting surfaces in which at least one of these is curved is a property of:
    Solution

    Lens is a transparent material bound by two surfaces, of which one or both surfaces are spherical.

    It may have two spherical surfaces, bulging outwards called convex lens and can have two spherical surfaces, curved inwards called concave lens. It may also have a plane and a spherical surface as  in plano-convex lens.

  • Question 3
    1 / -0
    A concave mirror produces $$2$$ times magnified real image of an object placed at $$5 cm$$ in front of it. Where is image located?
    Solution
    Magnification, $$m = \dfrac {-v}{u}$$
    Given, magnification, $$m = -2$$
    $$-2 = \dfrac {-v}{-5 cm}$$
    Therefore, object distance, $$v = -2\times 5 cm = -10 cm$$
  • Question 4
    1 / -0
    Mark the correct statement:
    Solution
    The reflecting surface of a spherical mirror forms a part of a sphere. The centre of this sphere is called the centre of curvature of the spherical mirror. The Center of curvature can also be defined as the point in the centre of the sphere from which the mirror was sliced. It is represented by the letter $$C$$. Please note that the centre of curvature of the concave mirror lies outside the mirror's reflecting surface. The centre of curvature of a convex mirror behind the mirror.
    So, only option-(A) is correct
  • Question 5
    1 / -0
    A concave mirror produces $$10$$ times enlarged image (real) of an object placed $$1 cm$$ in front of it. Where is the image located?
    Solution
    Given: 
    $$m = -10$$; magnification
    $$u = -1$$; Object distance

    $$m = \dfrac {-v}{u}$$

    $$\Rightarrow -10 = \dfrac {-v}{-1 }$$
    Therefore, the image distance, 
    $$v = -10\times 1 $$
    $$v =  -10\ cm$$
  • Question 6
    1 / -0
    Magnification produced by plane mirror is $$+1$$. It means:
    Solution
    Magnification $$= +1$$ signifies that the image formed in a plane mirror is of same size as the object. Positive sign in the value of magnification signifies that image formed by a plane mirror is erect.
  • Question 7
    1 / -0
    The focal length of a spherical mirror of radius of curvature $$40 cm$$ is
    Solution
    We know,
    $$ Focal\ length\ (f) = \dfrac{Radius\ of\ curvature\ (R)}{2} $$

    Given, 
    Radius of curvature, $$ R = 40\ cm $$
    $$ \Rightarrow f = \dfrac{R}{2} = \dfrac{40}{2} = 20\ cm $$

    Hence, the correct answer is OPTION B.
  • Question 8
    1 / -0
    A ray of light travels from air to glass at an angle of incidence of $$37^{\circ}$$, the angle of refraction is $$24^{\circ}$$. What is the refractive index of glass?
    (Given: $$\sin 37^{\circ} = 0.60, \sin 24^{\circ} = 0.40)$$
    Solution
    Snell's law states that ratio of sine of angle of incidence to sine of angle of refraction is a constant. This constant is called the refractive index of the medium. Hence:
    $${ n }_{ ga } = \dfrac {\sin r}{\sin r} = \dfrac {\sin 37^{\circ}}{\sin 24^{\circ}}$$
    $$= \dfrac {0.60}{0.40} = 1.5$$
  • Question 9
    1 / -0
    An object is placed at a distance of $$100 cm$$ from a concave lens of focal length $$40 cm$$. Find the distance of the position of image from the lens in $$cm$$.
    Solution
    By using sign convention rules
    Object distance, $$ u = -100\ cm $$
    Focal length of concave lens, $$ f = -40\ cm $$

    Using lens formula,
    $$ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $$
    We have, 

    $$ \Rightarrow \dfrac{1}{-40} = \dfrac{1}{v} - \dfrac{1}{-100} $$

    $$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{-40} - \dfrac{1}{100} $$

    $$ \Rightarrow \dfrac{1}{v} = \dfrac{-100-40}{4000} = \dfrac{-140}{4000} = \dfrac{-7}{200}$$

    $$ \Rightarrow v = \dfrac{-200}{7} = -28.57\ cm $$
    Magnification .

    $$m=\frac{v}{u}$$ = $$ \frac{-28.57}{-100}=0.2857$$
    so magnification positive means virtual image formed.


    "- " Indicates virtual image. 

    A virtual image is formed at a distance of $$ 28.57\ cm $$ from the optical center on the same side of the object. 

    Hence, the correct answer is OPTION B.
  • Question 10
    1 / -0
    A convex lens of focal length $$5\ cm$$ is placed at a distance of $$6\ cm$$ from a wall. How far from the lens should an object be placed so as to form its real image on the wall?
    Solution
    Given:
    Focal length of lens,$$f = 5cm$$
    Distance between lens and screen = distance of image formed from the lens$$=v = 6cm$$
    We know, $$\dfrac {1}{v} - \dfrac {1}{u} = \dfrac {1}{f}$$
    $$ \dfrac {1}{6} - \dfrac {1}{u} = \dfrac {1}{5}$$
    $$\dfrac {1}{6} - \dfrac {1}{5} = \dfrac {1}{u}$$
    $$\dfrac {5-6}{30} = \dfrac {1}{u}$$
    $$\dfrac{1}{u} = \dfrac {-1}{30}$$
    Therefore, the distance of the object from the lens, $$u = -30 cm$$. Or, the object should be placed $$30cm$$ far from the lens on the opposite side of the wall.
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