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ludmilkaskok [199]
3 years ago
6

Light illuminates two closely spaced thin slits and produces an interference pattern on a screen behind. How will the distance b

etween the fringes of the pattern differ for red light and blue light? 1. The same spacing for both 2. Closer for red light 3. Farther apart for red light
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

3. Farther apart for red light

Explanation:

The distance y between fringes is

y = \dfrac{\lambda L }{d},

where \lambda is the wavelength of light, L is the distance to the screen, and d is the slit separation.

Now, the wavelength \lambda_r of the red light is greater than for the wavelength \lambda_b for blue light:

\lambda_r> \lambda_b,

which means

\dfrac{\lambda_r L }{d}>\dfrac{\lambda_b L }{d},

In other words, the distance between the fringes is greater for red light, which from the options is choice 3.

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Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

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  • \rm k = 1.3\times 10^{-13} cm.
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The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

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\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

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