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Current Question (ID: 17551)

Question:

$\text{On an optical bench a point object is placed at the mark of 10 cm, a convex lens of focal length 15 cm at the mark of 40 cm and a concave lens of focal length 15 cm placed at the mark of 60 cm. The final image is formed at the mark of: (point object and two lenses are coaxial)}$

Options:

1. $30 \text{ cm}$
2. $80 \text{ cm}$
3. $90 \text{ cm}$
4. $\text{infinity}$

Solution:

$\text{Using lens formula:} \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\text{For convex lens:} \frac{1}{v_1} - \frac{1}{-30} = \frac{1}{15}$ $v_1 = 10 \text{ cm}$ $\text{Image formed at 50 cm acts as object for concave lens}$ $\frac{1}{v_2} - \frac{1}{-10} = -\frac{1}{15}$ $v_2 = 30 \text{ cm}$ $\text{Final image at 90 cm}$

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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
}