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Question
A 50 cm tall object is at a very large distance from a diverging lens. A virtual, erect and diminished image of the object is formed at a distance of 20 cm in front of the lens. How much is the focal length of the lens?
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Solution
Given: Height of object (ho) = 50 cm
Image distance (v) = −20 cm ...(Negative because it is virtual)
Object distance (u) = ?
By using the lens formula:
`1/f = 1/v + 1/u`
⇒ `1/f = 1/-20 + 1/u` ...(i)
Since the object is at a very large distance, we can assume u = ∞. Thus:
`1/u` = 0
Substituting this values in equation (i), we get,
⇒ `1/f = 1/-20 + 0`
⇒ `1/f = 1/-20`
⇒ f = −20
∴ The focal length of the lens is −20 cm.
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