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Current Question (ID: 17576)

Question:
$\text{The focal lengths of the objective and eyepiece of a compound microscope are } 2 \text{ cm and } 6.25 \text{ cm respectively. An object } AB \text{ is placed at a distance of } 2.5 \text{ cm from the objective which forms the image } A'B' \text{ as shown in the figure. The maximum magnifying power in this case, will be:}$
Options:
  • 1. $10$
  • 2. $20$
  • 3. $5$
  • 4. $25$
Solution:
$\text{Hint: } M = -\frac{v_0}{u_0} \left(1 + \frac{D}{f_e}\right)$ $\text{Step: Find the maximum magnifying power in this case.}$ $\text{The microscope is in distinct vision mode.}$ $\text{For objective-}$ $v = \frac{uf}{u+f} = \frac{(-2.5)(2)}{-2.5+2} = 10 \text{ cm}$ $M = -\frac{v_0}{u_0} \left(1 + \frac{D}{f_e}\right)$ $M = -\frac{10}{2.5} \left(1 + \frac{25}{6.25}\right) = 20$ $\text{Hence, option (2) is the correct answer.}$

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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
}