1. Plot the ray diagram for a spherical mirror for magnification m=-2/3
Answers
Answered by
0
Answer:
Like all problems in physics, begin by the identification of the known information.
ho = 4.0 cm
do = 35.5 cm
f = -12.2 cm
Next identify the unknown quantities that you wish to solve for.
di = ???
hi = ???
To determine the image distance (di), the mirror equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.
1/f = 1/do + 1/di
1/(-12.2 cm) = 1/(35.5 cm) + 1/di
-0.0820 cm-1 = 0.0282 cm-1 + 1/di
-0.110 cm-1 = 1/di
di = -9.08 cm
The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.
To determine the image height (hi), the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.
hi/ho = - di/do
hi /(4.0 cm) = - (-9.08 cm)/(35.5 cm)
hi = - (4.0 cm) • (-9.08 cm)/(35.5 cm)
hi = 1.02 cm
The negative values for image distance indicate that the image is located behind the mirror. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always indicates the existence of a virtual image located behind the mirror. In the case of the image height, a positive value indicates an upright image. Further information about the sign conventions for the variables in the Mirror Equation and the Magnification Equation can be found in Lesson 3.
From the calculations in this problem it can be concluded that if a 4.0-cm tall object is placed 35.5 cm from a convex mirror having a focal length of -12.2 cm, then the image will be upright, 1.02-cm tall and located 9.08 cm behind the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. Convex mirrors always produce images that are upright, virtual, reduced in size, and located behind the mirror
hope it helps
Similar questions