A DIVERGING lens has a focal length of 35 cm. (a) Find the image distance when an object is placed 48 cm from the lens. (b) Is the image real or virtual magnified or reduced?
To find the image distance and determine whether the image is real or virtual, magnified or reduced, we can use the lens formula:
1/f = 1/v - 1/u,
where f is the focal length of the lens, v is the image distance, and u is the object distance.
(a)
Given:
Focal length, f = 35 cm,
Object distance, u = 48 cm.
We can use the lens formula to find the image distance, v.
1/35 = 1/v - 1/48.
To solve for v, we can rearrange the equation as follows:
1/v = 1/35 + 1/48.
To add the fractions, we need to find a common denominator. In this case, the least common multiple of 35 and 48 is 840.
Multiplying both sides of the equation by 840, we get:
840/v = 840/35 + 840/48.
Simplifying the equation:
840/v = 24 + 17.5.
840/v = 41.5.
Cross-multiplying:
840 = 41.5v.
Dividing both sides of the equation by 41.5, we find:
v = 840/41.5.
Calculating this, we get:
v ≈ 20.241 cm.
Therefore, the image distance is approximately 20.241 cm.
(b)
To determine whether the image is real or virtual, magnified or reduced, we'll use the lens equation:
1/v - 1/u = 1/f.
Substituting the given values:
1/20.241 - 1/48 = 1/35.
Simplifying:
0.0494 - 0.0208 = 0.0286.
0.0286 ≠ 0.
Since the result is not equal to zero, the image formed by the diverging lens is virtual.
To determine if the image is magnified or reduced, we need to compare the image distance (v) to the object distance (u).
Here, the image distance (v) is less than the object distance (u), indicating that the image formed is reduced in size.
Therefore, the image formed by the diverging lens is virtual and reduced.