Question
-An object whose height is 5.8 cm is at a distance of 9.5 cm from a spherical concave mirror. Its image is real and has a height of 9.2 cm. Calculate the radius of curvature of the mirror.
I got 0.1165 m for my answer.
-How far from the mirror is it necessary to place the above object in order to have a virtual image with a height of 9.2 cm?
I know that I have two unknowns to solve for, but how do I go about solving them?
I got 0.1165 m for my answer.
-How far from the mirror is it necessary to place the above object in order to have a virtual image with a height of 9.2 cm?
I know that I have two unknowns to solve for, but how do I go about solving them?
Answers
Use the desired image height and the object height to get the magnification ratio: it is 9.2/5.8 = 1.586
Since you want the image to be virtual, the Do value must be negative, and such that |Di/Do| = 1.58 (the magnification)
Therefore Di = -1.58 Do
Use that together with the focusing equation
1/Do + 1/Di = 2/R
to solve for Do. You already know R.
Since you want the image to be virtual, the Do value must be negative, and such that |Di/Do| = 1.58 (the magnification)
Therefore Di = -1.58 Do
Use that together with the focusing equation
1/Do + 1/Di = 2/R
to solve for Do. You already know R.
I get as far as 1/Do + 1/-1.58 = 17.167, but I'm not sure what my next step is.
Wrong equation
1/Do + 1/-1.58Do = 17.167
Solve for Do
1/Do + 1/-1.58Do = 17.167
Solve for Do
Related Questions
A object of height 50cm is placed 100cm in front of a plane mirror.find
(.A)the image distance.
B...
When an object is droped from a certain height , it's distance from the starting point varies as the...
The gravitational pull between two objects depends on their mass and distance. What is meant by dist...
The gravitational pull between two objects depends on their mass and distance. What is meant by dist...