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Light Reflection and Refraction Test - 34

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Light Reflection and Refraction Test - 34
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  • Question 1
    1 / -0
    A ray of light travelling in air is incident on the plane of a transparent medium. The angle of incident is $$45^0$$ and that of refraction is $$30^0$$. The refractive index of the medium with respect to air is:
    Solution
    Given:
    Angle of incidence $$\angle i=45^0$$
    Angle of refraction $$\angle r=30^0$$
    Refractive index of transparent medium w.r.t. air is
    $$\mu_{ma}=\dfrac{\mu_m}{\mu_a}=\dfrac{\mu_m}{1}=\mu_m$$
    By Snell's Law,
    $$\displaystyle \mu_m=\dfrac {sin \ i}{sin \ r}=\dfrac {sin \ 45^0}{sin \ 30^0}=\cfrac {\dfrac {1}{\sqrt 2}}{\dfrac {1}{2}}=\sqrt 2$$
  • Question 2
    1 / -0
    Find the refractive index of glass with respect to water. The refractive indices of glass and air with respect to air are $$3/2$$ and $$4/3$$ respectively.
    Solution
    Refractive index of medium $$1$$ with respect to medium $$2$$, $$^1\mu_2=\dfrac {\mu_2}{\mu_1}$$
    Refractive index of glass with respect to air, $$^g\mu_a=\dfrac {3}{2}$$

    Refractive index of water with respect to air,$$^w\mu_a=\dfrac {4}{3}$$
    $$^g\mu_a=\dfrac {\mu^a}{\mu^g}$$

    $$^w\mu_a=\dfrac {\mu^a}{\mu^g}$$

    $$\therefore \text{refractive index of glass with respect to water,}\ \displaystyle ^g\mu_w=\dfrac {\mu_w}{\mu_a}\times \dfrac {\mu_a}{\mu_g}=\dfrac {^g\mu_a}{^w\mu_a}=\dfrac {3}{2}\times \dfrac {3}{4}=\dfrac {9}{8}$$
  • Question 3
    1 / -0
    In case of a virtual and erect image, the magnification of a mirror is :
    Solution
    For any erect image, the magnification of a mirror is positive.
    We know,
    Magnification (M) $$ = \dfrac{\text{height of image}(h_i)}{\text{height of object}(h_o)}$$ 
    The erect image formed same side of object to the principle axis. 
    So, In Sign convention, Sign of object height and image height will be same.
    Hence Magnification will be positive

  • Question 4
    1 / -0
    If the magnification of a body of size 1 m is 2, what is the size of the image?
    Solution
    $$m=2=\dfrac {h_i}{h_o}=\dfrac {h_i}{1}\Rightarrow h_1=2\times 1\ m$$
    So, $$h_1=2\ m$$
  • Question 5
    1 / -0
    Formula of focal length in convex lens is
    Solution
    $$\begin{array}{l}\text { Focal length of lens, from lens formula : } \\\qquad \begin{array}{l}\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\ \dfrac{1}{f}=\dfrac{u-v}{v \times u} \\ f=\dfrac{v \times u}{u-v}\end{array}\end{array}$$
  • Question 6
    1 / -0
    A light ray passing from crown glass to flint glass, then the speed of light is:
    Solution
    Refractive index of flint glass(1.65) is more than that of crown glass(1.55). So, the speed of light in flint glass is lesser than that in crown glass.
    Option B
  • Question 7
    1 / -0
    The rays parallel and close to the principal axis are called:
    Solution
    The rays parallel and close to the principal axis are called paraxial rays. The rays parallel but not close to the principal axis are called peripheral rays.

  • Question 8
    1 / -0
    The magnification produced by a concave mirror :
    Solution
    Depending on the position of the object the magnification produced by a concave mirror may be less than, equal to or greater than one.

  • Question 9
    1 / -0
    Ratio of the size of the image to the size of the object is known as:
    Solution
    Ratio of the size of the image to the size of the object is known as magnification. It is given by $$m = \dfrac{h_{i}}{h_{o}}$$
    where $$m \rightarrow$$  magnification
                $$h_{i} \rightarrow$$ height of image
                $$h_{O} \rightarrow$$ height of object
  • Question 10
    1 / -0
    The distance between the extreme points on the periphery of the mirror is called:
    Solution
    Principal section is also defined as the normal 'side view' of the mirror for a ray diagram. In the diagram AB is the principal section.

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