
A person can see clearly objects only when they lie between 50cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use will be
A. CONVEX, +0.15 D
B. Convex, +2.25 D
C. Concave, 0.25 D
D. Convex, +0.2 D
Answer
568.5k+ views
Hint: By using the mirror formula and power of the lens we can determine the type of lens . We are keeping the maximum distance so the person could see the object .
Complete step by step solution:
We need to accommodate a lens for a person to see the maximum distinct vision. Therefore the object distance is at ∞ and image distance is 400cm.
From the sign convention rule
\[u = - \infty \] To find the focal length for lens
\[
v = - 400cm \\
fe = ? \\
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} - \dfrac{1}{{ - \infty }} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} + \dfrac{1}{\infty } \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} - 0 \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} \\
f = - 400cm \\
\]
To find the power of the lens
\[
p = \dfrac{1}{f} \\
\Rightarrow p = \dfrac{1}{{ - 400cm}} \\
\Rightarrow p = \dfrac{1}{{ - 400 \times {{10}^{ - 2}}}} \\
\Rightarrow p = - 0.0025 \times {10^2} \\
\therefore p = - 0.25\,D \\ \]
Hence it is negative value; the lens used for the person is concave lens
Option (c) is correct
Note: Opticians prescribe correct lenses in indicating their power. If the lens prescribed has power (Positive value), then the lens prescribes convex. Similarly, the lens of power (negative value), then the lens prescribes concave. The power of the lens is the ability to bend the rays of Light. If the power of the lens is more bending of the light is more and lesser the focal length and vice versa.
Complete step by step solution:
We need to accommodate a lens for a person to see the maximum distinct vision. Therefore the object distance is at ∞ and image distance is 400cm.
From the sign convention rule
\[u = - \infty \] To find the focal length for lens
\[
v = - 400cm \\
fe = ? \\
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} - \dfrac{1}{{ - \infty }} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} + \dfrac{1}{\infty } \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} - 0 \\
\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{400}} \\
f = - 400cm \\
\]
To find the power of the lens
\[
p = \dfrac{1}{f} \\
\Rightarrow p = \dfrac{1}{{ - 400cm}} \\
\Rightarrow p = \dfrac{1}{{ - 400 \times {{10}^{ - 2}}}} \\
\Rightarrow p = - 0.0025 \times {10^2} \\
\therefore p = - 0.25\,D \\ \]
Hence it is negative value; the lens used for the person is concave lens
Option (c) is correct
Note: Opticians prescribe correct lenses in indicating their power. If the lens prescribed has power (Positive value), then the lens prescribes convex. Similarly, the lens of power (negative value), then the lens prescribes concave. The power of the lens is the ability to bend the rays of Light. If the power of the lens is more bending of the light is more and lesser the focal length and vice versa.
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