A farsighted person can see objects beyond 71 cm clearly if separation between glasses and eye lens is 2 cm, then find the focal length of the glass?
A. 23 cm
B. 34.5 cm
C. 18.4 cm
D. 30 cm

Hint:We have given that the person can see beyond 75 cm clearly that means for the object placed at distance of distinct vision (25 cm), the image should be formed at 75 cm from the eye so that the person can clearly see the object. Determine the object distance and image distance by subtracting the distance between the lens and eye from the image distance and object distance respectively. Use the lens formula to determine the focal length.

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

Complete step by step answer:
We have given that the person can see beyond 75 cm clearly that means for the object placed at distance of distinct vision (25 cm), the image should be formed at 75 cm from the eye. We have given the separation between glasses and eye lens is \[d = 2\,cm\]. Therefore, the object distance from the lens is,
\[u = - \left( {25 - 2} \right) = - 23\,{\text{cm}}\]
And the image distance is,
\[v = - \left( {71 - 2} \right) = - 69\,{\text{cm}}\]
We have the lens formula relating focal length, image distance and object distance,
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Substituting \[u = - 23\,{\text{cm}}\] and \[v = - 69\,{\text{cm}}\] in the above equation, we get,
\[\dfrac{1}{f} = \dfrac{1}{{ - 69}} - \dfrac{1}{{ - 23}}\]
\[ \Rightarrow \dfrac{1}{f} = - \dfrac{1}{{69}} + \dfrac{1}{{23}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{2}{{69}}\]
\[ \Rightarrow f = \dfrac{{69}}{2}\]
\[ \therefore f = 34.5\,{\text{cm}}\]
Since the focal length is positive, the lens that the person is using is the convex lens.

So, the correct answer is option B.

Note:Do not take the image distance as 71 cm since it is the distance between the eye lens and the image. The image distance and object distance should be taken from the pole of the lens. The distance of distinct vision is 25 cm and it is always negative.