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Wave Optics Test - 36

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Wave Optics Test - 36
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  • Question 1
    1 / -0
    How does the red shift confirm that the universe is expanding?
    Solution
    Red shift means shifting of colours in the spectrum of moving star towards the red end of the spectrum. It happens when the radiation emitting source goes away from the earth. It is due to Doppler's effect as we observe in case of sound. The apparent frequency decreases on wavelength increases for a recreding source. 
  • Question 2
    1 / -0
    Two sources of light of wavelengths 2500 $$\mathring {A} $$ and 3500 $$\mathring {A} $$ are used in Young's double slit experiment simultaneously. Which orders of fringes of two wavelength patterns coincide?
    Solution
    Let $$n^{th}$$ fringe of 2500$$\mathring {A} $$ coincide with $$(n -2)^{th}$$ fringe of 3500 $$\mathring {A} $$.
    $$\therefore 3500(n-2)=2500 \times n$$
    $$1000 n= 7000$$,
    $$n=7$$
    $$\therefore $$ 7th order fringe of 1 st source will coincide with 5th order fringe of 2nd source.
  • Question 3
    1 / -0
    An astronaut is looking down on earth's surface from a space shuttle an altitude of 400 km Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of visible light is 500 nm, the astronaut will be able to resolve linear objects of the size of about :
    Solution
    The resolving power of an instrument is given by the formula, $$R.P=1.22\times \frac{\lambda D}{d}$$
    Here, d is aperture of the instruments, D is distance of satellite from the earth. Here eye is the optical instruments.
    $$\displaystyle R.P=\frac{1.22\times 500\times 10^{-9}}{5\times 10^{-3}}\times 400\times 1000$$
              $$\displaystyle =1.22\times \frac{10^{-2}}{10^{-3}}\times 4 = 1.22\times 40=50 m$$
  • Question 4
    1 / -0
    In case of linearly polarised light, the magnitude of the electric field vector.
    Solution
    In any type of light whether polarised or unpolarised, the magnitude of electric field vector always varies periodically with time.
    Actually the change in electric field vector gives rise to periodically changing magnetic field.
  • Question 5
    1 / -0
    doctor prescribes spectacles to a patient with a combination of a convex lens of focal length 40 em, and concave lens of focal length 25 em then the power of spectacles will be.
    Solution
    For combination of lenses, power
    $$P=P_1+P_2=\frac {100}{40}-\frac{100}{25}=-1.5 D$$
  • Question 6
    1 / -0
    What change occurs, if the monochromatic light used in Young's double slit experiment is replaced by white light?
    Solution
    When we use white light in place of monochromatic light then only central fringe looks white and other fringes of different colour are observed. It is becaused central fringe for all the colours are formed at central point so it becomes white after mixiing up. As fringes of other colours fall at different places we see fringes of all colour on the central bright fringe.
  • Question 7
    1 / -0
    Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wave length of light =500 nm]
    Solution
    We have, $$\dfrac{y}{D}\ge 1.22 \dfrac{\lambda}{d}$$

    $$\Rightarrow D \le \dfrac{yd}{(1.22)\lambda} = \dfrac{10^{-3}\times 3\times 10^{-3}}{(1.22)\times 5\times 10^{-7}}$$

    $$=\dfrac{30}{6.1}\approx 5m$$

    $$\therefore  D_{max} = 5m$$
  • Question 8
    1 / -0
    Wavelength of light used in an optical instrument  are $$\lambda _1 = 4000 \mathring { A } $$ and $$ \lambda _2 = 5000 \mathring { A} $$ then ratio of their respective resolving powers(corresponding to $$\lambda _1$$ and $$ \lambda _2$$) is
    Solution
    We know, 
    $$ Resolving\ Power\ \propto \dfrac{1}{\lambda} $$

    Given, 
    $$ \lambda_1 = 4000 \mathring{A} $$
    $$ \lambda_2 = 5000 \mathring{A} $$

    $$ \Rightarrow \dfrac{RP_{\lambda_1}}{RP_{\lambda_2}} = \dfrac{ \lambda_2}{ \lambda_1} $$

    $$ \Rightarrow \dfrac{RP_{\lambda_1}}{RP_{\lambda_2}} = \dfrac{5000}{4000} = \dfrac{5}{4} $$

    $$ \Rightarrow RP_{\lambda_1} : RP_{\lambda_2} = 5:4 $$

    Hence, the correct answer is OPTION D. 
  • Question 9
    1 / -0
    Assertion: At the first glance, the top surface of the Morpho butterfly's wing appears a beautiful blue-green. If the wind moves the colour changes.
    Reason: Different pigments in the wing reflect light at different angles.
    Solution
    The assertion is true. When wind moves the colour ofthe wing changes. The visible colour of the wing is different from its original colour due to interference of light. Interference occurs between incident ray and reflected ray. Due to wind the reflectivity ofthe upper surface of the wing changes (as it is covered by tiny hairs) which changes the reflected wave. The resultant colour due to interference also changes. This is the principle behind the change of colour of wings of Morpho Butterfly. So, (c) is the answer.
  • Question 10
    1 / -0
    Assertion: Radio waves can'be polarised.
    Reason: Sound waves in air are longitudinal in nature.
    Solution
    Radio waves can be polarized because they are transverse in nature.
    And Sound waves in air are longitudinal in nature.
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