Answer:
cause you crazy..
Explanation:
The drawing shows a horizontal ray of white light incident perpendicularly on the vertical face of a prism made of crown glass. The ray enters the prism, and part of the light undergoes refraction at the slanted face and emerges into the surrounding material. The rest of the light is totally internally reflected and exits through the horizontal base of the prism. The colors of light that emerge from the slanted face of the prism may be chosen by altering the index of refraction of the material surrounding the prism. Find the required index of refraction of the surrounding material so that (a) only red light and (b) all colors except violet emerge from the slanted face of the prism. Take n
Answer:
The answer is "1.0748 and 1.0875".
Explanation:
Please find the complete question in the attachment file.
The incidence angle is [tex]i=45^{\circ}[/tex] for all colors When the angle is r, then use [tex]\frac{\sin{i}}{\sin{r}}=\frac{n_{o}}{n}[/tex] . Snell's rule Where [tex]n_{o}[/tex] is an outside material reflectance (same hue index) or n seems to be the crown glass index of the refraction, That index of inclination is [tex]90^{\circ}[/tex] as the light in color shifted behaver from complete inner diffraction to diffraction.
Whenever the external channel has a thermal conductivity for the red light, that's also
[tex]n_{o}=\frac{n_{r}\sin{45^{\circ}}}{\sin{90^{\circ}}}=\frac{1.520\times\sin{45^{\circ}}}{\sin{90^{\circ}}}=1.0748[/tex]
When outside the material has a refractive index, this happens with violet light.
[tex]n_{o}=\frac{n_{r}\sin{45^{\circ}}}{\sin{90^{\circ}}}=\frac{1.538\times\sin{45^{\circ}}}{\sin{90^{\circ}}}=1.0875[/tex]
In point a, The only red light flows out from the leaned face and the residual colors are mirrored mostly on prism for the primary benefits [tex]n_{o}=1.0748[/tex] (and slightly larger than that).
In point b, The only violet light is shown in the prism with the majority of the colors coming out from the sloping face for a scale similar to [tex]n_{o}= 1.0875[/tex] (and slightly smaller than this).
kid is bouncing on a pogo stick. he oscillates 22.0 times in 14.9 s. What is his period?
Answer:
Period = 0.68 seconds
Explanation:
Given the following data;
Number of oscillation = 22
Time = 14.9 seconds
To find the period;
Method I.
Period = time/number of oscillation
Period = 14.9/22
Period = 0.68 seconds.
Method II.
We would find the frequency of the wave;
Frequency = time/number of oscillation
Frequency = 22/14.9
Frequency = 1.48 Hertz
Next, we find the period;
Period = 1/frequency
Period = 1/1.48
Period = 0.68 seconds
A longitudinal wave is observed. Exactly 6 crests are observed
to move past a given point in 9.1 s. Its wavelength is 2.4 m and
its frequency is 0.66 HZ. What is the speed of the wave?