# One student uses a lens of focal length $50cm$ and another of $ - 50cm$. What is the nature of the lens and its power used by each of them?

Answer

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**Hint:**The distance of the focus from the surface of the lens or curved surface. The power of lens is defined as the ratio of inverse to the focal length of lens. Power of the lens unit is diopter(D). In the given question there are two students having their lens of focal length is given, from that we have to find out the power used by each student.

**Complete step-by-step solution:**

Given that,

One student lens of focal length is ${f_1} = 50cm$

And another student lens of focal length is ${f_2} = - 50cm$

The first student lens is convex lens because it has positive, whereas the second student lens is concave because it has negative,

Therefore, from the definition of power of lens, the focal length is inverse to the power of lens,

That means, ${P_1} = \dfrac{1}{{{f_1}}}$ it is the power of first student lens,

${P_2} = \dfrac{1}{{{f_2}}}$ is the power of a second student lens.

Therefore by substituting the values of focal length of two students, we get

${P_1} = \dfrac{1}{{50}}cm$ Centimeters is converted into meters that is,

$

{P_1} = \dfrac{{100}}{{50}} \\

{P_1} = 2D \\

$

Therefore for the second student is, ${P_2} = - \dfrac{1}{{50}}cm$

So that, $

{P_2} = - \dfrac{{100}}{{50}} \\

{P_2} = - 2D \\

$

Here the power is positive means it is convex lens and the power is negative means it is concave lens.

**Note:**Two students having lenses is the first student uses the convex lens because the power of the lens of the first student is positive. The second student uses the concave lens because the power of the lens of the second student is negative. Power is defined as the ability of a lens to converge or diverge a beam of light rays.

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