a converging mirror of focal length 20cm forms an image which is twice the size of the object .calculate 2 possible distances of the object from the mirror?
To solve this problem, we can use the mirror formula for a converging mirror:
1/f = 1/v - 1/u
where:
f is the focal length of the mirror,
v is the image distance,
and u is the object distance.
Given that the image is twice the size of the object, we can conclude that the magnification, m, is equal to 2. We know that the magnification is calculated as:
m = -v/u
Since m = -2 (as it's twice the size, it is a magnification of -2), we can substitute the value of m into the mirror formula:
1/f = 1/v - 1/u
-2 = 1/v - 1/u
Now, we need to find two possible distances for the object from the mirror.
Let's assume the first distance for the object is u1. Plugging in the values, we get:
-2 = 1/v1 - 1/u1
Similarly, let's assume the second distance for the object is u2:
-2 = 1/v2 - 1/u2
To find the possible distances for the object, we need to solve the above two equations.
Let's solve for u1:
-2 = 1/v1 - 1/u1
-2u1 = v1 - u1
-v1 = -2u1
v1 = 2u1
Let's solve for u2:
-2 = 1/v2 - 1/u2
-2u2 = v2 - u2
-v2 = -2u2
v2 = 2u2
Now, we have our two equations:
v1 = 2u1
v2 = 2u2
To find the two possible distances for the object, we can assume arbitrary values for v1 and v2, such as v1 = 40 cm and v2 = 30 cm.
Plugging these values into the equations, we get:
v1 = 40 cm = 2u1
40 = 2u1
u1 = 40/2
u1 = 20 cm
v2 = 30 cm = 2u2
30 = 2u2
u2 = 30/2
u2 = 15 cm
Therefore, the two possible distances for the object from the mirror are:
1) 20 cm
2) 15 cm