
A convex lens of focal length 30cm is combined with a concave lens of focal length 30cm. What is the power of the combined lens?
Answer
569.1k+ views
Hint: To find out the power of the combined lens, we need to find out the focal length of the combination. The power of the lens is given by the reciprocal of the focal length of the lens.
Formula used:
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
$\Rightarrow D = \dfrac{1}{f}$
Where, $f_1$, $f_2$, and f are the focal lengths of convex, concave and the combination of lenses respectively. D is the power of the combined lens.
Complete step by step answer
Let us consider, the focal length of the convex lens be $f_1$ and the focal length of the concave lens be $f_2$.
We know that, the combined focal length is given by the formula,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
We know that, the focal length of a convex lens is always positive and the focal length of the concave lens is always negative then,
$\Rightarrow {f_1} = 30cm$
and, ${f_2} = - 30cm$
So, when we substitute the values in the above formula, we get
$\Rightarrow \begin{array}{l}\dfrac{1}{f} = \dfrac{1}{{30}} - \dfrac{1}{{30}}\\ \Rightarrow \dfrac{1}{f} = \dfrac{0}{{30}}\end{array}$
Thus, we get the focal length of the combined lens as,
$\Rightarrow f = \dfrac{{30}}{0}cm = \infty$
Now, we know that the power of a lens is given by the reciprocal of the focal length of the lens. It is denoted by D.
Then,
$\Rightarrow D = \dfrac{1}{f}$
$\Rightarrow D = \dfrac{1}{\infty } = 0$
Thus, the power of the combination of the given convex and concave lens is 0.
Note
The focal length of a plane mirror is infinite and its power is equal to zero. Thus, the combination of convex and concave lens given in the question, works as a plane mirror. If the focal length of the convex lens is greater than the focal length of the concave lens, the combination works as a diverging lens, whereas, when the focal length of the convex lens is less than the focal length of the concave lens, the system works as a converging lens.
Formula used:
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
$\Rightarrow D = \dfrac{1}{f}$
Where, $f_1$, $f_2$, and f are the focal lengths of convex, concave and the combination of lenses respectively. D is the power of the combined lens.
Complete step by step answer
Let us consider, the focal length of the convex lens be $f_1$ and the focal length of the concave lens be $f_2$.
We know that, the combined focal length is given by the formula,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$
We know that, the focal length of a convex lens is always positive and the focal length of the concave lens is always negative then,
$\Rightarrow {f_1} = 30cm$
and, ${f_2} = - 30cm$
So, when we substitute the values in the above formula, we get
$\Rightarrow \begin{array}{l}\dfrac{1}{f} = \dfrac{1}{{30}} - \dfrac{1}{{30}}\\ \Rightarrow \dfrac{1}{f} = \dfrac{0}{{30}}\end{array}$
Thus, we get the focal length of the combined lens as,
$\Rightarrow f = \dfrac{{30}}{0}cm = \infty$
Now, we know that the power of a lens is given by the reciprocal of the focal length of the lens. It is denoted by D.
Then,
$\Rightarrow D = \dfrac{1}{f}$
$\Rightarrow D = \dfrac{1}{\infty } = 0$
Thus, the power of the combination of the given convex and concave lens is 0.
Note
The focal length of a plane mirror is infinite and its power is equal to zero. Thus, the combination of convex and concave lens given in the question, works as a plane mirror. If the focal length of the convex lens is greater than the focal length of the concave lens, the combination works as a diverging lens, whereas, when the focal length of the convex lens is less than the focal length of the concave lens, the system works as a converging lens.
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