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Current Question (ID: 17580)

Question:

$\text{The focal length of the objective lens and the eye lens is } 4 \text{ mm and } 25 \text{ mm}$ $\text{respectively in a compound microscope. The distance between the objective}$ $\text{and eyepiece lens is } 16 \text{ cm. Its magnifying power for relaxed eye position is:}$

Options:

1. $32.75$
2. $327.5$
3. $0.3275$
4. $\text{None of the above}$

Solution:

$\text{Hint: } m = \frac{(L_t - f_o - f_e)D}{f_o f_e}$ $\text{Step: Find the magnifying power for a relaxed eye position.}$ $f_o = 4 \text{ mm} = 0.4 \text{ cm}$ $f_e = 25 \text{ mm} = 2.5 \text{ cm}$ $\text{The length of the tube, } L_t = 16 \text{ cm}$ $\text{The magnifying power for relaxed eye position is given by;}$ $m = \frac{(L_t - f_o - f_e)D}{f_o f_e}$ $m = \frac{(16 - 0.4 - 2.5) \times 25}{0.4 \times 2.5}$ $\Rightarrow m = 327.5$ $\text{Hence, option (2) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}