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Current Question (ID: 17562)

Question:

$\text{The angle of incidence for a ray of light at a refracting surface of a prism is } 45^\circ . \text{ The angle of the prism is } 60^\circ . \text{ If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:}$

Options:

1. $45^\circ, \sqrt{2}$
2. $30^\circ, \sqrt{2}$
3. $30^\circ, \frac{1}{\sqrt{2}}$
4. $45^\circ, \frac{1}{\sqrt{2}}$

Solution:

$\text{Given: Angle of incidence=} 45^\circ, \text{ Angle of Prism=} 60^\circ$ $\text{Let us consider that, a light ray PQ incidences on surface AB and after passing through the prism, it moves along RS as shown in the figure:}$ $\text{Since the incident ray suffers minimum deviation, so it is clear that the ray inside the prism i.e. QR must be parallel to the base of the prism i.e. BC.}$ $\text{From above figure, we get following information-}$ $\text{Angle of refraction, } r = r' = 30^\circ$ $\text{Angle of emergence, } e = 45^\circ$ $\text{Minimum deviation suffered by the ray can be calculated using the formula as given below:}$ $\delta_{\min} = (i + e) - (r + r') = 90^\circ - 60^\circ = 30^\circ$ $\text{Also we know that}$ $\mu = \frac{\sin \left( \frac{A + \delta_{\min}}{2} \right)}{\sin \frac{A}{2}}$ $\text{A=angle of prism = } 60^\circ$ $\therefore \mu = \frac{\sin \left( \frac{60^\circ + 30^\circ}{2} \right)}{\sin \frac{60^\circ}{2}}$ $= \frac{\sin 45^\circ}{\sin 30^\circ}$ $= \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

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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
}