Import Question JSON

Current Question (ID: 17577)

Question:

$\text{The magnification of a compound microscope for the final image at the least distance of distinct vision is } 90. \text{ The magnification of the objective lens is } 15. \text{ The value of the focal length of the eyepiece will be:}$

Options:

1. $5 \text{ cm}$
2. $6 \text{ cm}$
3. $\frac{1}{6} \text{ cm}$
4. $12 \text{ cm}$

Solution:

$\text{For the least distance case, magnification of the eyepiece } = 1 + \frac{D}{f}$ $\text{Find the focal length of the eyepiece.}$ $m_e = 1 + \frac{D}{f}$ $\text{The magnification of the objective lens is } 15.$ $m = m_0 m_e$ $90 = 15 \left(1 + \frac{D}{f}\right)$ $\frac{D}{f} = 5$ $f = \frac{25}{5} = 5 \text{ cm}$ $\text{Therefore, the focal length of the eyepiece is } 5 \text{ cm.}$ $\text{Hence, option (1) is the correct answer.}$

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
}