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Current Question (ID: 17588)

Question:

$\text{An astronomical telescope has an objective and an eyepiece of focal lengths } 40 \text{ cm and } 4 \text{ cm respectively. To view an object } 200 \text{ cm away from the objective, the lenses must be separated by a distance:}$

Options:

1. $46.0 \text{ cm}$
2. $50.0 \text{ cm}$
3. $54.0 \text{ cm}$
4. $37.3 \text{ cm}$

Solution:

$\text{Hint: } \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\text{Step: Find the lenses that must be separated by a distance.}$ $\text{The focal length of the objective lens } (F_0) = +40 \text{ cm}$ $\text{The focal length of the eyepiece lens } (F_e) = +4 \text{ cm}$ $\text{The distance of the object from the objective lens } (U_0) = -200 \text{ cm}$ $\text{Applying the lens formula for the objective lens:}$ $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\Rightarrow \frac{1}{v} - \frac{1}{-200} = \frac{1}{40}$ $\Rightarrow \frac{1}{v} = \frac{1}{40} - \frac{1}{200} = \frac{5-1}{200} = \frac{4}{200}$ $\Rightarrow v = 50 \text{ cm}$ $\text{The formation of the image will occur at the } F_1 \text{ position i.e. first focus of the eyepiece lens.}$ $\text{The distance between the objective and the eyepiece lens for normal adjustment } d = v + F_e \Rightarrow 50 + 4 \Rightarrow 54 \text{ cm}$ $\text{Hence, option (3) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}