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# Light Rays and Optics

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Solve the following problems in the space provided. Show your work. Write equations in symbolic form before making numerical substitutions, and include the correct units with any numerical result.

1. Light of 520 nm wavelength is used in a double-slit (Young's) experiment (slit separation 0.45 m). Adjacent dark fringes separated by 2.10 mm are observed on a screen. Determine the slit-to-screen distance.

2. The image of a tree just covers the length of a 5.0-cm plane mirror when the mirror is held vertically 30.0 cm from the eye. The tree is 100 m from the mirror. What is its height?

3. Two thin lenses, the one on the left of focal length -0.200 m and the second of focal length of +0.200 m, are separated by 0.200 m. An object is 0.200 m to the left of the lens. Find the location of the final image both by a ray diagram and by a calculation using the thin-lens equation.

4. An object is located 10.0 cm from the center of a spherical silvered ball 5.0 cm in radius.

(a) Calculate the position of the image.
(b) What is the lateral magnification?
(c) Make a graphical solution.

5. You are given a diffraction grating having an unknown number of lines (rulings) per meter. Using the green line of mercury (546 x 10^-9 m), you find that the angle between the central and the first principal (most in¬tense) maxima is 30° How many lines/meter does the grating have?

https://brainmass.com/physics/system-work/light-rays-optics-152263

## SOLUTION This solution is FREE courtesy of BrainMass!

Hi there

Here are the solutions. Note that problem #2 is actually solved last.
As always they are in two file formats.

Note that the lenses and spherical mirror figures are drawn to scale and indeed agree with the numerical answer.

1. Light of 520 nm wavelength is used in a double-slit (Young's) experiment (slit separation 0.45 mm). Adjacent dark fringes separated by 2.10 mm are observed on a screen. Determine the slit-to-screen distance.

To get a constructive superposition on the screen (maximal intensity), the pathe difference  for the light rays coming from the slits must be an integer number of wavelengths:

But also:

Where d is the distance between the slits.

Hence:

If the distance D of the screen from the slits is very large compared to the slits width, we can write:

Now, the position of the maxima is:

But for small angles:

Therefore, the maxima of order m will appear at height y above the axis where y is given by:

Thus the spacing between two maximas is:

Since exactly between two maximas there is a dark fringe, the above expression is also the spacing between the dark pringes.

We can also see this from the requirement that for a dark fringe, the path difference is half a wavelength:

In our case we are looking for D:

3. Two thin lenses, the one on the left of focal length -0.200 m and the second of focal length of +0.200 m, are separated by 0.200 m. An object is 0.200 m to the left of the lens. Find the location of the final image both by a ray diagram and by a calculation using the thin-lens equation.

The intermediary image (the image from the left lens) is the red vertical line, while the final image of the system is the blue vertical line. The actual light path is denoted by the black arrows. The red dashed lines represent the ray tracing due to the left lens and the blue dashed lines are the ray tracing of the right lens.

Ray tracing:

For a concave lens (negative focal point), one ray begins parallel to the lens and reflected to the focal point. A second ray goes through the center of the lens. At the point of intersection an image is formed.

For a convex lens (positive focal point), one ray begins parallel to the lens and is refracted through the focal point on the other side. The second ray goes through the focal point and then refracted parallel to the axis. The image is formed where the rays meet (the second ray can also go uninterrupted through the center of the lens).

The lens equation is:

Where u is the object position with respect to the lens, v is the image location with respect to the lens and f is the focal distance.

Note that the distances in the lens formula are measured with respect to the center of each lens. An arrow that points to the left represents a negative value (for example, the first lens focal distance and image), while a right-pointing arrow represents a positive value.

We first solve for the concave lens, and use the formed image as the object for the second lens.

For the image of the first lens we get:

Note that as expected we get an image left of the lens.

Now, for the convex lens, note that the object is now formed 0.3m to the left of the lens:

So the final image is formed 0.6m to the right of the convex lens.

4. An object is located 10.0 cm from the center of a spherical silvered ball 5.0 cm in radius.

(a) Calculate the position of the image.
(b) What is the lateral magnification?
(c) Make a graphical solution.

For a convex mirror the focal length is half the radius of curvature.

Ray tracing:

One ray begins parallel to the axis and reflected in such a way that its imaginary continuation into the mirror passes through the focal point.
The second ray goes to the intersection point of the mirror and the axis and is reflected with the same angle with respect to the axis. The intersection of the two imaginary continuations of the rays defines the location of the image.

Like the thin lenses the equation for spherical mirror is:

For convex mirror, the focal point length and the image location are negative since they are both inside the mirror.

Thus in our case:

This means that the image is 2 centimeters into the mirror.

The magnification is defined as:

Which in our case yields:

This means that the image will be upright and 1/5 the size of the object.

5. You are given a diffraction grating having an unknown number of lines (rulings) per meter. Using the green line of mercury (546 x 10^-9 m), you find that the angle between the central and the first principal (most in¬tense) maxima is 30° How many lines/meter does the grating have?

The condition for a maxima in a diffraction grating is the same as in the two slits experiments, namely that the path difference of two adjacent rays will be an integer multiple of the wavelength.

We have already seen in the first problem that for a spacing d we have:

At the center m=0, so the diffraction angle with respect to the axis for order n is simply:

Or simply:

In our case:

So the distance between two slits is 1.1m, so the spacing in units of slits per meter is:

2. The image of a tree just covers the length of a 5.0-cm plane mirror when the mirror is held vertically 30.0 cm from the eye. The tree is 100m from the mirror. What is its height?

The ray-tracing for the tree-mirror-eye is shown below:

The eye is located at point O, the mirror is between points B and E while the tree is between points A and G. Note that the line OH does not necessarily intersects the center of the mirror.

The distance from the tree to the mirror is D, while the distance from the eye to the mirror is d

Also, FE is parallel to OH and to CB.

Hence, due to the reflection rules:

Or in other words triangle EHO is similar to triangle EFG and triangle OHB is similar to triangle CBA

Thus,

And together:

The height of the tree is therefore:

In our case this gives:

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