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Current Question (ID: 17517)

Question:

$\text{The slab of a refractive index material equal to } 2 \text{ shown in the figure has a curved surface } APB \text{ of a radius of curvature of } 10 \text{ cm and a plane surface } CD. \text{ On the left of } APB \text{ is air and on the right of } CD \text{ is water with refractive indices as given in the figure. An object } O \text{ is placed at a distance of } 15 \text{ cm from the pole } P \text{ as shown. The distance of the final image of } O \text{ from } P \text{ as viewed from the left is:}$

Options:

1. $20 \text{ cm}$
2. $30 \text{ cm}$
3. $40 \text{ cm}$
4. $50 \text{ cm}$

Solution:

$\text{Hint: } \frac{\mu_1}{v} - \frac{\mu_2}{u} = \frac{\mu_1 - \mu_2}{R}$ $\text{Step: Find the distance of the final image of } O \text{ from } P$ $\text{From the formula of the refraction from the curve surface for an object kept at a distance } u = -15 \text{ cm and radius of curvature } R = -10 \text{ cm, then,}$ $\frac{\mu_1}{v} - \frac{\mu_2}{u} = \frac{\mu_1 - \mu_2}{R}$ $\frac{1}{v} - \frac{2}{-15} = \frac{1 - 2}{-10}$ $\Rightarrow v = -30 \text{ cm}$ $\text{It will be a virtual image there will be no refraction at the plane surface } CD, \text{ so the final distance of the image will be } 30 \text{ cm.}$ $\text{The curved surface will form a virtual image } l \text{ at a distance of } 30 \text{ cm from } P. \text{ Since the image is virtual there will be no refraction at the plane surface } CD, \text{ (as the rays are not actually passing through the boundary), the distance of the final image } l \text{ from } P \text{ will remain } 30 \text{ cm.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}