NEET-XII-Physics
17: Light Waves
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- Qstn #35A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution if it is known to be between 1.2 and 1.5?Ans : Given:
Thickness of soap film, d =0.0011 mm = 0.0011 × 10-3 m
Wavelength of light used, `` \lambda =580\,\mathrm{\,nm\,}=580\times {10}^{-9}\,\mathrm{\,m\,}``
Let the index of refraction of the soap solution be μ.
The condition of minimum reflection of light is 2μd = nλ,
where n is an interger = 1 , 2 , 3 ...
`` \Rightarrow \,\mathrm{\,\mu \,}=\frac{n\lambda }{2d}=\frac{2n\lambda }{4d}``
`` =\frac{580\times {10}^{-9}\times \left(2n\right)}{4\times 11\times {10}^{-7}}``
`` =\frac{5.8\left(2n\right)}{44}=0.132\left(2n\right)``
As per the question, μ has a value between 1.2 and 1.5. So,
`` n=5``
`` \,\mathrm{\,So\,},\,\mathrm{\,\mu \,}=0.132\times 10=1.32``
Therefore, the index of refraction of the soap solution is 1.32.
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- Qstn #36A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index = 1.4). What should be the minimum thickness of the film so that it strongly reflects the light?Ans : Given:
Wavelength of light used, `` \,\mathrm{\,\lambda \,}=560\times {10}^{-9}\,\mathrm{\,m\,}``
Refractive index of the oil film, `` \,\mathrm{\,\mu \,}=1.4``
Let the thickness of the film for strong reflection be t.
The condition for strong reflection is
`` 2\,\mathrm{\,\mu \,}t=\left(2n+1\right)\frac{\lambda }{2}``
`` \Rightarrow t=\left(2n+1\right)\frac{\lambda }{4\,\mathrm{\,\mu \,}}``
where n is an integer.
For minimum thickness, putting n = 0, we get:
`` t=\frac{\lambda }{4\,\mathrm{\,\mu \,}}``
`` =\frac{560\times {10}^{-9}}{4\times 1.4}``
`` ={10}^{-7}\,\mathrm{\,m\,}=100\,\mathrm{\,nm\,}``
Therefore, the minimum thickness of the oil film so that it strongly reflects the light is 100 nm.
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- Qstn #37A parallel beam of white light is incident normally on a water film 1.0 × 10-4 cm thick. Find the wavelengths in the visible range (400 nm - 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33.Ans : Given,
Wavelength of light used, `` \,\mathrm{\,\lambda \,}=400\times {10}^{-9}\,\mathrm{\,m\,}\,\mathrm{\,to\,}700\times {10}^{-9}\,\mathrm{\,nm\,}``
Refractive index of water, `` \,\mathrm{\,\mu \,}=1.33``
The thickness of film, `` t={10}^{-4}\,\mathrm{\,cm\,}={10}^{-6}\,\mathrm{\,m\,}``
The condition for strong transmission: `` 2\,\mathrm{\,\mu \,}t=n\lambda ``,
where n is an integer.
`` \Rightarrow \lambda =\frac{2\,\mathrm{\,\mu \,}t}{n}``
`` \Rightarrow \lambda =\frac{2\times 1.33\times {10}^{-6}}{n}``
`` =\frac{2660\times {10}^{-9}}{n}\,\mathrm{\,m\,}``
Putting n = 4, we get, λ1 = 665 nm.
Putting n = 5, we get, λ2 = 532 nm.
Putting n = 6, we get, λ3 = 443 nm.
Therefore, the wavelength (in visible region) which are strongly transmitted by the film are 665 nm, 532nm and 443 nm.
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- Qstn #38A glass surface is coated by an oil film of uniform thickness 1.00 × 10-4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm - 750 nm) which are completely transmitted by the oil film under normal incidence.Ans : Given:
Wavelength of light used, `` \lambda =400\times {10}^{-9}\,\mathrm{\,to\,}750\times {10}^{-9}\,\mathrm{\,m\,}``
Refractive index of oil, μoil, is 1.25 and that of glass, μg, is 1.50.
The thickness of the oil film, `` d=1\times {10}^{-4}\,\mathrm{\,cm\,}={10}^{-6}\,\mathrm{\,m\,},``
The condition for the wavelengths which can be completely transmitted through the oil film is given by
`` \,\mathrm{\,\lambda \,}=\frac{2\,\mathrm{\,\mu \,}d}{\left(n+{\displaystyle \frac{1}{2}}\right)}``
`` =\frac{2\times {10}^{-6}\times \left(1.25\right)\times 2}{\left(2n+1\right)}``
`` =\frac{5\times {10}^{-6}}{\left(2n+1\right)}\,\mathrm{\,m\,}``
`` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{5000}{\left(2n+1\right)}\,\mathrm{\,nm\,}``
Where n is an integer.
For wavelength to be in visible region i.e (400 nm to 750 nm)
When n = 3, we get,
`` \lambda =\frac{5000}{\left(2\times 3+1\right)}``
`` =\frac{5000}{7}=714.3\,\mathrm{\,nm\,}``
When, n = 4, we get,
`` \lambda =\frac{5000}{\left(2\times 4+1\right)}``
`` =\frac{5000}{9}=555.6\,\mathrm{\,nm\,}``
When, n = 5, we get,
`` \lambda =\frac{5000}{\left(2\times 5+1\right)}``
`` =\frac{5000}{11}=454.5\,\mathrm{\,nm\,}``
Thus the wavelengths of light in the visible region (400 nm - 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455 nm.
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- Qstn #39Plane microwaves are incident on a long slit of width 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30°.Ans : Given:
Width of the slit, b = 5.0 cm
First diffraction minimum is formed at θ = 30°.
For the diffraction minima, we have:
bsinθ = nλ
For the first minima, we put n = 1.
`` 5\times \,\mathrm{\,sin\,}30°=1\times \,\mathrm{\,\lambda \,}``
`` \Rightarrow \lambda =\frac{5}{2}=2.5\,\mathrm{\,cm\,}``
Therefore, the wavelength of the microwaves is 2.5 cm.
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- Qstn #40Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and falls on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?Ans : Given:
Wavelength of the light used, `` \,\mathrm{\,\lambda \,}=560\,\mathrm{\,nm\,}=560\times {10}^{-9}\,\mathrm{\,m\,}``
Diameter of the pinhole, d = 0.20 mm = 2 × 10-4 m
Distance of the wall, D = 2m
We know that the radius of the central bright spot is given by
`` R=1.22\frac{\lambda D}{d}``
`` =1.22\times \frac{560\times {10}^{-9}\times 2}{2\times {10}^{-4}}``
`` =6.832\times {10}^{-3}\,\mathrm{\,m\,}\,\mathrm{\,or\,}=0.683\,\mathrm{\,cm\,}``
Hence, the diameter 2R of the central bright spot on the wall is 1.37 cm.
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- Qstn #41A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light is focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?Ans : Given:
Wavelength of the light used, `` \,\mathrm{\,\lambda \,}=620\,\mathrm{\,nm\,}=620\times {10}^{-9}\,\mathrm{\,m\,}``
`` ``
Diameter of the convex lens, `` d=8\,\mathrm{\,cm\,}=8\times {10}^{-2}\,\mathrm{\,m\,}``
Distance from the lens where light is to be focused, `` D=20\,\mathrm{\,cm\,}=20\times {10}^{-2}\,\mathrm{\,m\,}``
The radius of the central bright spot is given by
`` R=1.22\frac{\lambda D}{d}``
`` =1.22\times \frac{620\times {10}^{-9}\times 20\times {10}^{-2}}{8\times {10}^{-2}}``
`` =1891\times {10}^{-9}\,\mathrm{\,m\,}\approx 1.9\times {10}^{-6}\,\mathrm{\,m\,}``
∴ Diameter of the central bright spot, 2R = 3.8 × 10-6 m.