Question:
If a person cannot see objects closer than 100 cm, the power of lens required to read at 25 cm is:
If a person cannot see objects closer than 100 cm, the power of lens required to read at 25 cm is:
Show Hint
Hypermetropia always requires a convex (converging) lens to help focus light closer, meaning your final answer must have a positive power value (\( + \)). This allows you to immediately eliminate negative choices during exams.
Updated On: Jun 8, 2026
- +3D
- -3D
- +4D
- -4D
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
The person is suffering from hypermetropia (farsightedness) since their near point has shifted out to 100 cm instead of the standard 25 cm. To fix this, a corrective lens must create a virtual image of an object placed at the standard reading distance (\( u = -25~\text{cm} \)) at the person's actual defective near point (\( v = -100~\text{cm} \)).
We calculate this using the classical lens formula:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
The power of the lens in diopters is given by \( P = \frac{100}{f\text{ (in cm)}} \).
Step 1: Substituting coordinate variables into the lens formula.
Using sign convention:
• Object distance, \( u = -25~\text{cm} \)
• Image distance, \( v = -100~\text{cm} \)
\[ \frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} = -\frac{1}{100} + \frac{1}{25} \]
Step 2: Solving for the focal length \( f \).
\[ \frac{1}{f} = \frac{-1 + 4}{100} = \frac{3}{100} \implies f = \frac{100}{3}~\text{cm} \]
Step 3: Calculating the lens power \( P \).
\[ P = \frac{100}{f} = \frac{100}{\frac{100}{3}} = +3~\text{D} \] Thus, a converging lens with a power of \( +3~\text{D} \) is required.
Step 1: Substituting coordinate variables into the lens formula.
Using sign convention:
• Object distance, \( u = -25~\text{cm} \)
• Image distance, \( v = -100~\text{cm} \)
\[ \frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} = -\frac{1}{100} + \frac{1}{25} \]
Step 2: Solving for the focal length \( f \).
\[ \frac{1}{f} = \frac{-1 + 4}{100} = \frac{3}{100} \implies f = \frac{100}{3}~\text{cm} \]
Step 3: Calculating the lens power \( P \).
\[ P = \frac{100}{f} = \frac{100}{\frac{100}{3}} = +3~\text{D} \] Thus, a converging lens with a power of \( +3~\text{D} \) is required.
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Optics Questions
- A thin oil layer floats on water. A ray of light making an angle of incidence \( 45^\circ \) shines on the oil layer. The angle of refraction of light ray in water is:
- The resultant amplitude of superposition of two waves \( y_1 = a \cos(\omega t) \) and \( y_2 = a \cos(\omega t + \phi) \) is:
- The image formed by a convex mirror of focal length 30 cm is a quarter of the size of the object. The distance of the object from the mirror is
- A diffraction pattern is obtained by using a beam of red light. If the red light is replaced by the blue light, then
- The critical angle for diamond with respect to air is nearly
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions