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?
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(a) To find the image distance, we can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length of the lens (given as 35 cm)
v = image distance (what we are trying to find)
u = object distance (given as 48 cm)
First, let's substitute the given values into the lens formula:
1/35 = 1/v - 1/48
Let's simplify this equation. To do that, we need to find a common denominator for v and 48. The simplest way is to multiply both sides of the equation by 35v and 48v:
48v - 35v = 35(48) - 35(35)
13v = 1680 - 1225
13v = 455
Now divide both sides of the equation by 13:
v = 35
Therefore, the image distance is 35 cm.
(b) To determine if the image is real or virtual, and whether it is magnified or reduced, we need to examine the signs of the distances:
- If the image distance (v) is positive, the image is real.
- If the image distance (v) is negative, the image is virtual.
In this case, v is positive (35 cm), so the image is real.
To determine if the image is magnified or reduced, we can use the magnification formula:
magnification (m) = -v/u
where:
m = magnification
v = image distance (35 cm)
u = object distance (48 cm)
Now let's substitute the values:
m = -35/48
The magnification is negative, which indicates that the image is inverted compared to the object. The magnitude of the magnification (|m|) tells us if the image is magnified or reduced.
In this case, the magnification is less than 1 (|m| < 1), so the image is reduced.
Therefore, in summary:
(a) The image distance is 35 cm.
(b) The image is real and reduced.