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Ray Optics and Optical Test - 72

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Ray Optics and Optical Test - 72
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  • Question 1
    1 / -0
    A thin glass (refractive index 1.5) lens has optical power of -5D in air.Its optical power in a liquid medium with refractive index 1.6 will be: 
    Solution

  • Question 2
    1 / -0
    A thin Prism $$P_1$$ with angle $$4^o$$ and made from glass of refractive index 1.54 is combined with another thin prism $$P_2$$ made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of Prism $$P_2$$ is 
    Solution

  • Question 3
    1 / -0
    A beam of the light incident vertically on a glass hemisphere of radius R and refractive index $$\sqrt 2$$ lying with its plane side on a table .The axis of beam coincides with the vertical axis passing through the center of the base of the hemisphere and the cross sectional radius is $$\cfrac{R}{\sqrt2}$$ .The luminious spot form on the table is 
    Solution

  • Question 4
    1 / -0
    Inside a 6 cm thick glass slab a rectangular cavity is made as shown. On one face of slab a cross is marked. On other side an observes is looking i=on the slab. Find the shift of cross as seen by the man

    Solution
    Shift produced-
    $$S = \left( {1 - \dfrac{1}{\mu }} \right) \times \left( {l - d} \right)$$
    $$l = 6{\text{ cm}}$$
    $$d = 2{\text{ cm}}$$
    $$\mu  = \dfrac{3}{2}$$
    So,
    $$S = \left( {1 - \dfrac{2}{3}} \right) \times \left( {6 - 2} \right){\text{ cm}}$$
    $$S = \left( {\dfrac{1}{3} \times 4} \right){\text{ cm}} = \dfrac{4}{3}{\text{ cm}}$$

  • Question 5
    1 / -0
    For path $$A \rightarrow B$$ optical path is

    Solution
    $$AC = \dfrac{a}{\cos \, 60^o} ; CB = \dfrac{b}{\cos 30^o}$$
    $$\sin 60^o = \mu \sin 30^o$$
    $$\mu = \sqrt{3}$$
    optical path $$= AC + \mu CB$$
    $$= \dfrac{a}{cos 60^o} + \mu \dfrac{\beta}{cos \, 30^o} = 2a + \sqrt{3} \dfrac{6}{\dfrac{\sqrt{3}}{2}} = 2 (a + b)$$

  • Question 6
    1 / -0
    Light propagates $$2\ cm$$ distance in glass of refractive index $$1.5$$ in time $$t_0$$. In the same time $$t_0$$, light propagates a distance of $$2.25\ cm$$ in a medium. The refractive index of the medium is :
    Solution
    In medium-1 (refractive index $$n_1=1.5$$), light travels a distance $$2\ cm$$ in time $$t_0$$.
    The speed of light in medium-1 is $$v_1 = \dfrac{2\times 10^{-2}\ m}{t_0\ sec}$$
    In medium-2 (refractive index $$n_2$$), light travels a distance $$2.25\ cm$$ in time $$t_0$$.
    The speed of light in medium-2 is $$v_2 = \dfrac{2.25\times 10^{-2}\ m}{t_0\ sec}$$
    From the definition of refractive index, we can write,
    $$\dfrac{n_1}{n_2}=\dfrac{v_2}{v_1}$$
    $$\Rightarrow \dfrac{1.5}{n_2}= \dfrac{2.25\times 10^{-2}}{2\times 10^{-2}}$$
    $$\Rightarrow n_2 = \dfrac{1.5\times 2\times 10^{-2}}{2\times 10^{-2}}= \dfrac{4}{3}$$
  • Question 7
    1 / -0

    Directions For Questions

    Consider the set up shown in the figure, which uses a right-angled prism to transmit a horizontal incident ray. This device is known as anamorphic de-magnifier. During its passage through the prism, the ray experience total internal reflection $$(TIR)$$ twice and then exits as shown in the figure. The prism in made from a glass of refractive index $$\mu$$. 

    ...view full instructions

    Range the values of tip angle $$\theta$$ for the advice to work as anamorphic de-magnifier is
    Solution

  • Question 8
    1 / -0
    The refractive index of material of a prism of angles $$45^{\circ}, -45^{\circ}$$, and $$-90^{\circ}$$ is $$1.5$$. The path of the ray of light incident normally on the hypotenuse side is shown in
  • Question 9
    1 / -0
    A light ray is incident ton face AB of prism ABC as shown in figure. The second prism is kept in such a manner that emergent ray from prism ABC is falling normally on face A'B' of prism A'B'C'. The net deviation produced by the optical' system of the two prisms is :

    Solution

  • Question 10
    1 / -0
    A ray enters a glass slab at an angle $$ \alpha $$ from air and the refractive index varies along the slab as $$ \mu = \mu_0 - kt^2 $$ where $$ t $$ is the distance measured along the normal . $$ \mu_0 $$ and $$ k $$ are positive constants . If the glass slab is sufficiently thick , how far along the normal will the ray go before  it reflects back ? 
    Solution

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