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Current Question (ID: 17545)

Question:

$\text{A lens having focal length } f \text{ and aperture of diameter } d \text{ forms an image of intensity } I. \text{ An aperture of diameter } \frac{d}{2} \text{ in central region of lens is covered by a black paper. The focal length of lens and intensity of the image now will be respectively:}$

Options:

1. $f \text{ and } \frac{I}{4}$
2. $\frac{3f}{4} \text{ and } \frac{I}{2}$
3. $f \text{ and } \frac{3I}{4}$
4. $\frac{f}{2} \text{ and } \frac{I}{2}$

Solution:

$\text{The focal length of the lens will remain the same when half of the aperture is covered with black paper}$ $\text{But the intensity of the image is directly proportional to the area.}$ $\text{Intensity } I \propto A^2$ $\Rightarrow \frac{I_2}{I_1} = \left[ \frac{A_2}{A_1} \right]^2 = \frac{\pi^2 - \frac{\pi^2}{4}}{\pi^2} = \frac{3}{4}$ $\Rightarrow I_2 = \frac{3}{4} I_1.$

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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
}