8. [5pt] What is the image distance? Answer: 10. [5pt] A red light is submerged
ID: 1514047 • Letter: 8
Question
8. [5pt]
What is the image distance?
Answer:
10. [5pt]
A red light is submerged 2.5 m beneath the surface of a liquid with an index of refraction 1.35. What is the radius of the circle from which light escapes from the liquid into the air above the surface?
Answer:
A 23cm tall object is placed in front of a concave mirror with a radius of 21cm. The distance of the object to the mirror is 78cm.
11. [5pt]
Calculate the focal length of the mirror.
Answer:
12. [5pt]
Calculate the image distance.
Answer:
13. [5pt]
Calculate the magnification of the image (Remember, a negative magnification
corresponds to an inverted image).
Answer:
14. [5pt]
Calculate the magnitude of the image height.
Answer:
Explanation / Answer
9) DACACBE
10) Radius = 2.5 * tan47.79 = 2.756 m
11) Here, focal length = 21/2 = 10.5 cm
12) 1/v = 1/10.5 -1/78
=> image distance = 12.13 cm
13) magnification of image = -v/u = -12.13/78
= - 0.155
14) magnitude of the image height = 23 * 0.155 = 3.565 cm
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.