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Light Reflection and Refraction Test - 56

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Light Reflection and Refraction Test - 56
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  • Question 1
    1 / -0
    The refractive index of water and glass with respect to air is $$1.3$$ and $$1.5$$ respectively. Then the refractive index of glass with respect to water is
    Solution
    As we know,
    $$_w\mu_g =\dfrac {_a\mu_g}{_a\mu_w}=\dfrac {1.5}{1.3}$$
  • Question 2
    1 / -0
    A wave has velocity v in medium P and velocity 2v in medium Q. If the wave is incident in medium P at an angle of $$30^o$$, then the angle of refraction will be 
    Solution
    Let the absolute refractive index of medium P and medium Q are $$n_p$$ and $$n_q$$ respectively.
    Given:
    The velocity of the wave in medium P is $$v$$
    The velocity of the wave in medium Q is $$2v$$
    Angle of incidence $$i=30^o$$ in medium P
    The angle of refraction $$r=?$$ in medium Q
    From the definition of the refractive index, we can write,
    $$n_p = \dfrac{c}{v}$$ and $$n_q = \dfrac{c}{2v}$$
    Thus, $$\dfrac{n_p}{n_q} = \dfrac{2v}{v} = 2$$
    Now, from Snell's law we know, $$n_p sin\ i = n_q sin\ r$$
    $$\Rightarrow \dfrac{n_p}{n_q} = \dfrac{sin\ r}{sin\ i}$$
    $$\Rightarrow 2=\dfrac{sin\ r}{sin\ 30^o}$$
    $$\Rightarrow sin\ r = 2\times \dfrac{1}{2}=1$$
    $$\Rightarrow sin\ r= sin\ 90^o$$
    $$\therefore r=90^o$$
  • Question 3
    1 / -0
    A light moves from denser to rarer medium, which of the following is correct?
    Solution
    Let refractive index of denser medium $$=n_1$$
    refractive index of rarer medium $$=n_2$$
    We know,
    $$n_{21} = \dfrac{n_2}{n_1} = \dfrac{speed\ of\ light\ in\ denser\ medium}{speed\ of\ light \ in\ rarer \ medium}$$

    $$\dfrac{n_2}{n_1} = \dfrac{v_1}{v_2}$$

    $$n_1 > n_2$$

    $$\implies v_2 > v_1$$

    $$\therefore$$ velocity increases.
    The frequency of light does not change on traveling from one medium to another.
  • Question 4
    1 / -0
    What happens when light ray passes through a glass slab?
    Solution
    When a light ray passes from rarer to denser medium then light ray bends towards normal and when a light ray passes from denser medium to rarer medium then it bends away from normal.
  • Question 5
    1 / -0
    Which of the following signifies Snell's law?
    Solution
    Snell's law state that the ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. 
    $$ \dfrac {sin \ i}{ sin\ r}= constant $$

  • Question 6
    1 / -0
    When a light ray, incident at an angle passes through a glass slab then _______ ray is shifted laterally. 
    Solution
    When a light ray, incident at an angle passes through a glass slab then emergent ray is shifted laterally
  • Question 7
    1 / -0
    A monochromatic ray of light enters a glass slab $$\left( n=1.5 \right) $$ along the normal to the surface. The angle of deviation of the refracted ray is:
    Solution
    Ray of light enters the glass slab normally.
    Angle of incidence, $$i=0^0$$
    Let the angle of refraction be $$r$$.

    Using Snell's law: 
    $$n_1\sin\ i=n_2\sin \ r$$
    $$1\ \sin\ 0^0=n_2\sin \ r$$
    $$\sin \ r=0^0$$
    $$\Rightarrow r=0^0$$

    This means that the ray continues along its original path implying no deviation.
  • Question 8
    1 / -0
    Identify which of the following mirrors converge the parallel light rays to its focus ?
    Solution
    A concave mirror is a converging mirror. 
    The focus of a concave mirror is the point on the principal axis where incident parallel rays meet after reflection.
  • Question 9
    1 / -0
    A light ray parallel with the principle axis falls on a thin lens having a focal point $$f$$ as shown in the above figure. Identify which of the following statements best describes the lens?

    Solution
    The rays parallel to the principal axis diverge after passing through the lens. This implies that the lens must be diverging lens i.e concave lens.
    Hence the lens is thicker at the edges and thinner at the center.
  • Question 10
    1 / -0
    According to Snell's Law of refraction, the ratio of the refractive index is related to the inverse ratio of the ____ of the angles.
    Solution
    According to Snell's law of refraction :     $$n_1 \times sin i  = n_2 \times sin r$$
    $$\implies$$  $$\dfrac{n_1}{n_2} = \dfrac{sin r }{sin i}$$
    Thus ratio of indices is related to the inverse ratio of the sine of angles.
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