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Current Question (ID: 17499)

Question:

$\text{The speed of light in media } M_1 \text{ and } M_2 \text{ is } 1.5 \times 10^8 \text{ m/s and } 2.0 \times 10^8 \text{ m/s respectively. A ray of light enters from medium } M_1 \text{ to } M_2 \text{ at an incidence angle } i. \text{ If the ray suffers total internal reflection, the value of } i \text{ is:}$

Options:

1. $\text{equal to } \sin^{-1}\left(\frac{2}{3}\right)$
2. $\text{equal to or less than } \sin^{-1}\left(\frac{3}{5}\right)$
3. $\text{equal to or greater than } \sin^{-1}\left(\frac{3}{4}\right)$
4. $\text{less than } \sin^{-1}\left(\frac{2}{3}\right)$

Solution:

$\text{Hint: For total internal reflection, } \sin i \geq \sin \theta_c$ $\text{Step: Find the value of the incidence angle } i.$ $\text{In total internal reflection, the angle of incidence } (i) \text{ must be greater than the critical angle } (\theta_c)$ $\text{The refractive index for the medium } M_1 \text{ is given by:}$ $\mu_1 = \frac{c}{v_1} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$ $\text{The refractive index for the medium } M_2 \text{ is given by:}$ $\mu_2 = \frac{c}{v_2} = \frac{3 \times 10^8}{2.0 \times 10^8} = \frac{3}{2}$ $\mu_1 = 2 \text{ and } \mu_2 = \frac{3}{2}$ $2 \sin i \geq \frac{3}{2} \sin 90^\circ$ $\Rightarrow \sin i \geq \frac{3}{4}$ $\Rightarrow i \geq \sin^{-1}\left(\frac{3}{4}\right)$ $\text{So, when the angle of incidence is greater than or equal to the critical angle } \left[ \sin^{-1}\left(\frac{3}{4}\right) \right], \text{ the total internal reflection will happen.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}