Question

A person is unable to see objects nearer than 50cm. He wants to read a book placed at a distance of 25cm. Find the nature of focal length and power of the lens, he needs for his spectacles.

Answer 1

Hint: If this person wants to read the book placed at 25cm, the image of the book should be formed at 50cm so as he can read the book. Use the lens equation to determine the focal length of the lens. The power of the lens is reciprocal of the focal length.

Formula used:
Lens equation, \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\] ,
where, f is the focal length, u is the object distance and v is the image distance.
Power, \[P = \dfrac{1}{f}\],
where, f is the focal length.

Complete step by step answer:
We can see the person is suffering from the Hypermetropia also known as farsightedness. Therefore, the image of the object forms beyond the retina of the eye with Hypermetropia. So, we need a convex lens to converge the rays onto the retina.We have given that the person can see the objects placed beyond 50cm. If this person wants to read the book placed at 25cm, the image of the book should be formed at 50cm so he can read the book. Since both image and object are on the same side of the lens, the sign of both object and image is negative.We have the lens formula,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Here, f is the focal length, u is the object distance and v is the image distance.

Substituting \[v = - 50\,{\text{cm}}\] and \[u = - 25\,{\text{cm}}\] in the above equation, we get,
\[\dfrac{1}{f} = \dfrac{1}{{ - 50}} - \dfrac{1}{{ - 25}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 50}} + \dfrac{2}{{50}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{50}}\]
\[ \Rightarrow f = 50\,{\text{cm}}\]
Therefore, the focal length of the spectacles is 50 cm or \[0.5\,{\text{m}}\].
Now, we know that the power of the lens is reciprocal of the focal length. Therefore,
\[P = \dfrac{1}{f}\]
Substituting \[f = 0.5\,{\text{m}}\]in the above equation, we get,
\[P = \dfrac{1}{{0.5}}\]
\[ \therefore P = + 2\,{\text{D}}\]

Thus, the power of the convex lens of the spectacles should be \[ + 2\,{\text{D}}\].

Note:Students must note that 50cm is not the object distance but the image distance for which the person can see the objects if the image of the object forms at 50cm or beyond 50cm. One should be able to recognize the nature of the lens required to cure farsightedness and nearsightedness. Also, one can recognize the nature of the lens by looking at the sign of the focal length. If the focal length is positive, the lens must be a convex lens. Make sure that you have converted the focal length in units of meters.