An object is placed 11 cm in front of a convex mirror of focal length 4 cm. Using the lens equation, find where the image will form and state whether it is a real or virtual image.
Let x be the unknown distance of the image from the mirror.
Solve
1/11 + 1/x = 1/f = 1/4
x will be positive so the image will be real.
To find where the image will form and determine whether it is real or virtual, we can use the lens equation:
1/f = 1/v - 1/u
Where:
- f is the focal length of the mirror
- v is the image distance or the distance at which the image is formed
- u is the object distance or the distance of the object from the mirror
Given:
- f = 4 cm
- u = -11 cm
Now, let's substitute the values into the lens equation and solve for v:
1/4 = 1/v - 1/-11
To simplify, we'll take the reciprocal of -11:
1/4 = 1/v + 1/11
Now, we'll find a common denominator and add the fractions:
1/4 = (11 + v) / (11v)
Multiplying both sides of the equation by 11v:
11v/4 = 11 + v
Expanding:
11v = 44 + 4v
Subtracting 4v from both sides:
11v - 4v = 44
7v = 44
Dividing both sides by 7:
v = 44/7
So, v = 6.29 cm.
Now, to determine whether the image formed is real or virtual, we need to consider the sign convention. In this convention:
- Positive object distance (u) is measured from the object towards the mirror (to the left)
- Positive image distance (v) is measured from the image towards the mirror (to the left)
- If the image distance (v) is negative, the image is virtual
- If the image distance (v) is positive, the image is real
In this case, since v = 6.29 cm, which is positive, the image formed is real.
Therefore, the image will form at 6.29 cm from the convex mirror and it will be a real image.