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Ray Optics and Optical Test - 58

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Ray Optics and Optical Test - 58
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  • Question 1
    1 / -0
    The image formed by a convex mirror is only one-third of the size of the object. If the focal length of the mirror is $$12 \,cm,$$ the image formed will be
    Solution
    Given, 
    Height of image $$(h_1)$$ =$$ \dfrac{height \,of \,object (h_0)}{3}$$       Focal length of mirror (f) = 12 cm

    Magnification = $$m = \dfrac{h_f}{h_o} = \dfrac{-v}{u} = \dfrac{1}{3} \Rightarrow u = -3v$$

    Using Mirror formula ,$$\Rightarrow \dfrac{1}{v} + \dfrac{1}{u}  = \dfrac{1}{f}$$

    $$\Rightarrow \dfrac{1}{v} + \dfrac{1}{-3v} = \dfrac{1}{12}$$

    $$\Rightarrow \dfrac{2}{3v} = \dfrac{1}{12}$$

    $$\Rightarrow v = 8 \,cm$$             As v is positive, so image is formed at $$8 \,cm$$ behind the mirror.
  • Question 2
    1 / -0
    When a ray of light from air enters a denser medium, it:
    Solution

    The ray of light bends towards the normal.
    $$\text{Reason:}$$ As the speed of light decreases in the denser medium, it bends towards the normal.

  • Question 3
    1 / -0
    A ray of light moving from an optically rarer to a denser medium 
    Solution
    A ray of light moving from an optically rarer to a denser medium then bends towards the normal. While when a ray of light moves from denser to rarer then it bends away from the normal.
  • Question 4
    1 / -0

    A $$10 \mathrm{mm}$$ long awl pin is placed vertically in front of a concave mirror. A $$5 \mathrm{mm}$$ long image of the awl pin is formed at $$30 \mathrm{cm}$$ in front of the mirror. The focal length of this mirror is:

    Solution

     Given: object size $$\mathrm{h}=10
    \mathrm{mm},$$ image size $$\mathrm{h}^{\prime}=5 \mathrm{mm},$$ image distance
    $$\mathrm{v}=-30 \mathrm{cm},$$

    To find object  distance=?  $$\mathrm{f}=?$$

    from the magnification formula $$=\dfrac{h^{\prime}}{h}=\dfrac{-v}{u}$$

    or $$\dfrac{5
    m m}{10 m m}=\dfrac{-30 c m}{u}$$ or $$\dfrac{-30 c m}{u}=\dfrac{1}{2}$$

    or $$u=-60$$

    Now, it can be calculated as follows:

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

    or $$\dfrac{1}{-30
    \mathrm{cm}}+\dfrac{1}{-60 \mathrm{cm}}=\dfrac{1}{f}$$ or $$\dfrac{-2-1}{60}=\dfrac{1}{f}$$

    $$\dfrac{1}{f}=-\dfrac{1}{20
    \mathrm{cm}}$$ or $$f=-20 \mathrm{cm}$$  (the negative sign confirms that the calculated focal length is of a concavemirror)

  • Question 5
    1 / -0
    A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from :
    Solution
    Principal or first focus is the point from which incident rays after passing through convex lens, become parallel to the principal axis.
  • Question 6
    1 / -0
    A thin convex lens of focal length $$10\ cm$$ is placed in contact with a concave lens of same material and of same focal length. he focal length of combination will be
    Solution
    As we know,
    $$f=\dfrac{f_1f_2}{f_1+f_2}=\dfrac{10(-10)}{10+(-100)}=\dfrac{-100}{10-10}=\infty$$
  • Question 7
    1 / -0
    An object is placed at a distance of $$20 \ cm$$ from a convex lens of focal length $$10\ cm$$. The image is formed on the other side of the lens at a distance
    Solution
    As we know,
    $$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$ (Given $$u=-20\ cm,\ f=10\ cm,\ v=?$$)
    $$\therefore \dfrac{1}{10}=\dfrac{1}{v}-\dfrac{1}{(-20)} \Rightarrow v=20\ cm$$
  • Question 8
    1 / -0
    A convex lens makes a real image $$4\ cm$$ long on a screen. When the lens is shifted to a new position without disturbing the object, we again get a real image on the screen which is $$16\ cm$$ tall. The length of the object must be
    Solution
    As we know,
    $$O=\sqrt{I_1I_2}=\sqrt{4 \times 16}=8\ cm$$
  • Question 9
    1 / -0
    The refractive indices of glass and water w.r.t air are $$\dfrac{3}{2}$$ and $$\dfrac{4}{3}$$ respectively. The refractive index of glass w.r.t water will be
    Solution
    As we know,
    $$_a\mu_{g}=\dfrac {3}{2},  _a\mu_{w}=\dfrac {4}{3}$$
    $$\therefore \ _w\mu_{g}=\dfrac {_a\mu_{g}}{_a\mu_{w}}=\dfrac {3/2}{4/3}=\dfrac 98$$
  • Question 10
    1 / -0
    A beam of light is converging towards a point $$I$$ on a screen. A plane glass plate whose thickness in the direction of the beam $$=t$$ refractive index $$=\mu$$ is introduced in the path of the beam. The convergence point is shifted by
    Solution
    Normal
    shift $$\Delta x=\left(1-\dfrac 1 \mu \right)t$$
    and shift takes place in direction of ray.

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