A 0.49 cm high object is placed 8.5 cm in front of a diverging lens whose focal length is -7.5 cm. What is the height of the image?
To find the height of the image formed by a lens, we can use the lens formula:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens
- v is the distance of the image from the lens
- u is the distance of the object from the lens
In this case, the object distance (u) is given as 8.5 cm and the focal length (f) is given as -7.5 cm (since it's a diverging lens). We have to be careful with the signs: if the lens is diverging, the focal length is negative.
We need to find the image height (h').
To relate the object height (h) and the image height (h') formed by a lens, we can use the magnification formula:
h'/h = -v/u
Where:
- h is the object height
- h' is the image height
- v is the distance of the image from the lens
- u is the distance of the object from the lens
In this case, the object height (h) is given as 0.49 cm.
Now, let's calculate the image distance, v, using the lens formula:
1/f = 1/v - 1/u
Substituting the given values, we get:
1/-7.5 = 1/v - 1/8.5
Simplifying the equation, we have:
-1/7.5 = 1/v - 1/8.5
To simplify the equation further, we can take the least common denominator of 7.5 and 8.5, which is 63.75:
-1/7.5 = (8.5 - v)/v * 63.75
Simplifying again, we get:
-1/7.5 = (8.5 - v)/v * 63.75
To solve for v, we can cross-multiply and solve the resulting equation:
-1 * v * 63.75 = (8.5 - v) * 7.5
-63.75v = 63.75 - 7.5v
Combine like terms:
-56.25v = 63.75
Divide both sides by -56.25:
v = 63.75 / -56.25
v ≈ -1.13 cm
Now, we can use the magnification formula to find the image height (h'):
h'/h = -v/u
Substituting the known values:
h' / 0.49 = -(-1.13) / 8.5
Simplifying:
h' / 0.49 = 1.13 / 8.5
Cross-multiplying:
h' * 8.5 = 0.49 * 1.13
h' ≈ 0.067
Therefore, the height of the image formed by the diverging lens is approximately 0.067 cm.