I'm working on a final physics exam, and I've got everything but these two that I'm stuck on. Any help would be appreciated, and explanations too (:
--A------B---(-----
A and B are both pointing up.
1. ^ In that diagram, the image of object A would be
A) real, reduced, and upright.
B) virtual, enlarged, and upright
C) virtual, reduced, and inverted.
D) virtual, reduced, and upright.
and
2. Monochromatic light shines on the surface of a diffraction grating with 5.3 x 103 lines/cm. The first-order maximum is observed at an angle of 17°. Find the wavelength.
A) 420 nm
B) 520 nm
C) 530 nm
D) 550 nm
Thanks again!
1. To determine the image of object A in the given diagram, we need to analyze the ray diagrams for mirrors or lenses. In this case, it seems like we are dealing with mirrors.
In the diagram "A---B---(---)", A and B are both pointing up. This suggests that A and B are both mirror elements (such as concave or convex mirrors) that are facing each other. The "F" in the center indicates the focal point.
Now, based on the given options, let's analyze each one of them:
A) Real, reduced, and upright: A real image would be formed if the rays actually converge to a point in front of the mirror. Since A is pointing toward B, which is its mirror partner, they are not converging. Additionally, a reduced image would be smaller than the object, and an upright image would be the same way up as the object. Therefore, this option is incorrect.
B) Virtual, enlarged, and upright: A virtual image is formed when rays appear to diverge from a point behind the mirror. Since the arrows of A and B are pointing toward each other, it suggests that rays diverge and create a virtual image. Additionally, an enlarged image is larger than the object. However, since option A was incorrect, we can conclude that the image is not enlarged. Therefore, this option is also incorrect.
C) Virtual, reduced, and inverted: A virtual image, as explained earlier, is formed when rays appear to diverge. The image would be considered reduced if it appears smaller than the object, and inverted if it is upside down compared to the object. Since the arrows of A and B are pointing toward each other, suggesting rays diverge, and since the object is a certain way up and the image appears upside down, option C fits the description. This is the correct option.
D) Virtual, reduced, and upright: We have already discussed why this option is incorrect. The image is not upright.
Therefore, the answer to question 1 is C) virtual, reduced, and inverted.
2. To find the wavelength in a diffraction grating experiment, we can use the formula:
wavelength = distance between slits / number of slits
Given that the diffraction grating has 5.3 x 10^3 lines/cm, we need to convert it to lines/mm to match the unit of the distance between the slits. Therefore:
number of slits = 5.3 x 10^3 lines/cm * (1 cm / 10 mm)
number of slits = 5.3 x 10^2 lines/mm
Now, we have the number of slits as 5.3 x 10^2 lines/mm.
Considering the first-order maximum is observed at an angle of 17°, we can use the following formula:
sin(angle) = m * (wavelength / distance between slits)
For the first order (m = 1), we have:
sin(17°) = 1 * (wavelength / 5.3 x 10^2 lines/mm)
Rearranging the equation to solve for wavelength:
wavelength = sin(17°) * 5.3 x 10^2 lines/mm
Now we can calculate the value:
wavelength = sin(17°) * 5.3 x 10^2 lines/mm = 0.2923 * 5.3 x 10^2 lines/mm
After the calculation, we find:
wavelength ≈ 1.5469 x 10^2 lines/mm
Since the unit given in the options is nm (nanometers), we need to convert this value to nanometers by multiplying by 10:
wavelength ≈ 1.5469 x 10^2 lines/mm * 10 nm/mm = 1.5469 x 10^3 nm
Rounding to the nearest whole number, the value is approximately:
wavelength ≈ 1550 nm
Therefore, the answer to question 2 is D) 550 nm.
Good luck with your physics exam!