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?
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:
Where d is the distance between the slits.
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:
Which leads to:
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 ...
This solution contains annotated ray diagrams and step-by-step calculations to determine the slit-to-screen distance, height, location of final image, image position, lateral magnification, and angle between central and first principal maxima.