Question

A lens forms a vertical image 14 cm away from it when an object is placed 10 cm away from it. The lens is a....... less of focal length.......
A. concave, 6.67 cm
B. concave, 2.86 cm
C. convex, 2.86 cm
D. none

Answer 1

Hint: For a diverging lens, the distance of the object from the optical centre is equal to the distance of the image of the optical centre. And in this case of diverging lens both the distance of object and the distance of image from the optical centre is negative.

 Complete step by step answer:
According to the sign convention, the data are,
Distance of the object from the lens is $u = - 10\;{\rm{cm}}$.
Distance of the image from the lens is $v = - 14\;{\rm{cm}}$.
Express the lens formula to find the focal length of the lens.
\[\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}\]
Here, f is the focal length.
Substitute, $u = - 10\;{\rm{cm}}$, $v = - 14\;{\rm{cm}}$ to find the value of $f$.
\[\begin{array}{l}
\dfrac{1}{{ - 10\;{\rm{cm}}}} = \dfrac{1}{{ - 14\;{\rm{cm}}}} - \dfrac{1}{f}\\
\dfrac{1}{f} = \dfrac{1}{{ - 14\;{\rm{cm}}}} + \dfrac{1}{{10\;{\rm{cm}}}}\\
f = 35\;{\rm{cm}}
\end{array}\]
Since the focal length of the lens is positive, therefore, the lens is a convex lens.

So, the correct answer is “Option D”.

Note:
The distance between focal point and lens is called focal length. The focal length is always positive for converging lenses while the focal length is always negative for the converging length. For a diverging lens, the focal length is negative and it is the distance from the point where the combination of beams appear to be diverging after passing through the lens. The focal length is positive for diverging lenses.