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Salsk061 [2.6K]
3 years ago
13

Suppose that a pane of crown glass, with a refractive index of 1.52, is immersed in water, which has a refractive index of 1.33.

A ray of light traveling in the water strikes the glass at 45° relative to the normal and travels through the pane. What angle, relative to the normal, will the ray of light subtend when it leaves the pane and reenters the water?
Physics
1 answer:
mylen [45]3 years ago
3 0

Answer:

the angle made by this ray with normal will be 45°

Explanation:

given,                                                      

refractive index of crown glass = 1.52

refractive index of the water = 1.33        

ray of light in water strikes glass at 45°                      

angle subtended by the light when it reenter water = ?                  

When light enter in the glass from the water it get bend toward normal because refractive index of glass is more than water.              

And when ray comes out of the glass it is parallel to the initial light ray.

hence, the angle made by this ray with normal will be 45°

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An electron is trapped in an infinite square-well potential of width 0.6 nm. If the electron is initially in the n = 4 state, wh
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Answer:

E₁ = 1.042 eV

E₄₋₃= 7.29 eV      

E₄₋₂= 12.50 eV

E₄₋₁= 15.63 eV

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E₃₋₁= 8.34eV

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E = \frac {h^{2} n^{2}}{8mL^{2}}      

<em>where, h: Planck constant = 6.62x10⁻³⁴J.s, n: is the energy state, m: mass of the electron and L: widht of the square-well potential </em>      

<u>The energy of the electron in the ground state, </u><u>n = 1</u><u>, is:  </u>

E_{1} = \frac {(6.62 \cdot 10^{-34})^{2} (1)^{2}}{(8) (9.11 \cdot 10^{-31}) (0.6 \cdot 10^{-9} m)^{2}}    

E_{1} = 1.67 \cdot 10^{-19} J = 1.042 eV      

The photon energies that are emitted as the electron jumps to the ground state is the difference between the states:                      

E_{\Delta n} = \Delta n^{2} E_{1}  

E_{(4 - 3)} = (4^{2} - 3^{2}) 1.042 eV = 7.29eV

E_{(4 - 2)} = (4^{2} - 2^{2}) 1.042 eV = 12.50eV

E_{(4 - 1)} = (4^{2} - 1^{2}) 1.042 eV = 15.63eV

E_{(3 - 2)} = (3^{2} - 2^{2}) 1.042 eV = 5.21eV

E_{(3 - 1)} = (3^{2} - 1^{2}) 1.042 eV = 8.34eV

E_{(2 - 1)} = (2^{2} - 1^{2}) 1.042 eV = 3.13eV    

Have a nice day!                          

7 0
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