An object is placed on the principal axis of a lens of power -10 D. at a distance of 15 cm. The image formed is __________.

Correct option is C

Given:

Power of the lens is = −10  The lens is a diverging lens (since the power is negative).

Object Distance = 15cm

Formula used:

​$$\text{Lens formula:} \quad \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\$$ 

Solution:

Substitute the value:

​$$\frac{1}{-10} = \frac{1}{v} - \frac{1}{-15} \\$$ 

$$\frac{1}{-10} = \frac{1}{v} + \frac{1}{15} \\$$ 

$$\frac{1}{v} = \frac{1}{-10} - \frac{1}{15} \\$$ 

$$\frac{1}{v} = \frac{-3}{30} - \frac{2}{30} = \frac{-5}{30} \\$$ 

$$v = \frac{-30}{5} = -6 \, \text{cm} \\$$ 

$$\text{The image is formed at a distance of} \quad v = -6 \, \text{cm}, \text{which means the image is virtual, erect, and reduced in size.}$$ 

Thus the correct answer is (C)

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