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# Light

1. The so-called "hydrogen alpha (Ha) line" is a specific wavelength of red light caused by excited atomic hydrogen. It is frequently used when studying the Sun. The wavelength of Ha is 656.3 nm, or 656.3*10-9 m. The speed of light in vacuum is 3*108 m/s. What is the frequency ( in units of cycles per second, or 1/s) of Ha in vacuum?

2. Light is incident at angle A from a media with n = 1.5 to one with n = 2.0 as shown in the figure. What is the minimum value for A so that there is total internal reflection at point P, which is at the interface between the n = 2 and n = 1 materials?

3. An object of height 5 cm is placed 30 cm in front of spherical concave mirror. If the image is real and 10 cm high, what is the radius of curvature of the mirror?

4. The so-called "lensmakers equation" is shown below. The focal length of the lens (in cm) is f. The constant n is the index of refraction of the glass, relative to the medium around it (which is just the listed value of n unless the lens is submerged in water, or some other liquid). R1 and R2 are the radii of curvature of the two sides of the lens. The value of R for a side is considered to be positive if that side is convex; otherwise, it is negative. If the lens is flat on one side, the value of R for that side is infinite, and the ratio (1/R) is equal to zero.

Assume the lens is biconvex; that is, that both sides are convex. The radii of curvature are both 20 cm. The coefficient of refraction is 1.40. What is the focal length of the lens, in cm?

5. The index of refraction of a glass varies with the wavelength of the light passing through it. This has important consequences for the design of optical instruments. A type of glass known as "extra-dense flint" has an n = 1.7378 for blue light, and an n of 1.7130 for red light. Calculate the focal length of the lens above, for both red and blue light.

You will find that red and blue light form images at slightly different distances from the lens. This results in a blurred image. The problem is solved by using systems of lenses composed of glasses having different indices of refraction, and different radii of curvature. Such systems are called either achromatic or apochromatic, depending upon the degree of correction. Designing such a system is a non-trivial exercise, and is beyond the scope of this course.

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For this exercise, you will investigate the phenomenon of refraction. The simulation allows you to experiment with changes in the angles of incidence and combinations of materials. Do the following:

Click on the following link to access the simulation .

Fendt, W. (1997). Refraction of light (simulation). Retrieved on March 1, 2008, from http://www.walter-fendt.de/ph14e/refraction.htm

The first part of the simulation is a series of measurements designed to demonstrate the validity of Snell's Law. The light ray passes from medium 1, which has an index of refraction n1, into medium 2, with an index of refraction of n2. The angle of incidence is θ1, the angle of refraction θ2. Complete the table below. (The first line has been completed for you.) Explain in detail how your results demonstrate the validity of Snell's Law.

 Medium 1 n1 Medium 2 n2 target q1 actual q1 n1sinq1 q2 n2sinq2 Vacuum 1.00 Quartz 1.46 30 30.3 0.50 20.2 0.50 Vacuum Quartz 60 Vacuum Quartz 90 Vacuum SF2 30 Vacuum SF2 60 Vacuum SF2 90 Vacuum Diamond 30 Vacuum Diamond 60 Vacuum Diamond 90 BK7 SF2 30 BK7 SF2 60 BK7 SF2 90

The second part of the simulation examines total internal reflection. If you've ever snorkeled or SCUBA-dived, you've noticed this. If the water is smooth, and you look nearly straight up, you can see objects above the water. If you look beyond a certain angle, you see only water. That "certain angle" in the critical angle, θcrit.

In the simulation, adjust the angle of incidence until the angle of refraction is just equal to 90 degrees. Record the data in the table below. Explain your results in terms of Snell's Law.

 Medium 1 n1 Medium 2 n2 qcrit sinqcrit n2/n1` Water 1.33 Air 1.00 49.1 0.76 0.75 SF2 1.33 Diamond 1.65 Diamond 1.00

Thanks for asking me to reply these questions. I will look into these questions and let you know whether I can reply or not.
Customer: replied 5 years ago.

Ok. I am also willing to provide some compensation or bonus because I know the subscription might not be able to cover the cost.

Thanks.

It is day time here and I will try these problems today. Thanks.
I could solve all these problems and will try to upload the answers today itself.