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Ray Optics

282253 A light ray falls on a rectangular glass slab as shown. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is

1

\sqrt{3 / 2}

2

\frac{(\sqrt{3} + 1)}{2}

3

\frac{(\sqrt{2} + 1)}{2}

4

\sqrt{5} / 2

Explanation:

AGiven, $i = 45^{\circ}, r + θ_{c} = 90^{\circ}$ (from figure) $r =$ angle of refraction.
$θ_{c} =$ critical angle
Using Snell's law-
$μ = \frac{Sin i}{Sin r}$
$Sin r = \frac{Sin 45^{\circ}}{μ} = \frac{1}{\sqrt{2} μ}$
$Sin (90^{\circ} - θ_{c}) = \frac{1}{\sqrt{2} μ}$
$\frac{\sqrt{μ^{2} - 1}}{μ} = \frac{1}{\sqrt{2} μ}$
$\Rightarrow μ^{2} - 1 = \frac{1}{2}$
$\Rightarrow μ = \sqrt{3 / 2}$

AIIMS-27.05.2018(M)

Ray Optics

282254 An optical illusion which occurs mainly in deserts during hot summer is based on the principle of

1 Reflection

2 Interference

3 Dispersion

4 Total internal reflection

NDA (I) 2017

Ray Optics

282255 The critical angle for glass is $41^{\circ} 48^{'}$ and that for water is $48^{\circ} 36^{'}$ . Calculate the critical angle for glass-water interface.

1

62^{\circ} 43^{'}

2

34^{\circ} 42^{'}

3

52^{\circ} 42^{'}

4

44^{\circ} 42^{'}

Explanation:

A: Given that,
Critical angle for glass $(C_{g}) = 41^{\circ} 48^{'}$
Critical angle for water $(C_{w}) = 48^{\circ} 36^{'}$
Critical angle of glass water interface is
$\sin C_{g / w} = \frac{1}{μ_{g} / μ_{w}} = \frac{μ_{w}}{μ_{g}}$
For glass, $\sin C_{g} = \frac{1}{μ_{g}}$
For water $\sin C_{w} = \frac{1}{μ_{w}}$
$∴ \sin C_{g / w} = \frac{\sin C_{g}}{\sin C_{w}} = \frac{\sin {41.48}^{'}}{\sin {48.36}^{'}}$
Convert minute into degree. Minute is multiplied by $\frac{1}{60}$
$\begin{array}{r} = \frac{\sin {41.8}^{\circ}}{\sin {48.6}^{\circ}} = 0.89 \\ C_{g / w} = {62.87}^{\circ} ◻ 62^{\circ} 43^{'} \end{array}$

JCECE-2017

Ray Optics

282256 A beam of light from a source $L$ is incident normally on a plane mirror mixed at a certain distance $x$ from the source. The beam is reflected back as a spot on a scale placed just above the source $L$ . When the mirror is rotated through a small angle $θ$ , the spot of the light is found to move through a distance $y$ on the scale. The angle $θ$ is given by

1

\frac{y}{2 x}

2

\frac{y}{x}

3

\frac{x}{2 y}

4

\frac{x}{y}

Explanation:

A: According to question,

If mirror is rotated by angle $θ$ then angle of reflection will be rotated by $2 θ$ (because of law of reflection) Therefore, $\tan 2 θ = \frac{y}{x}$ $2 θ = \frac{y}{x}$ $θ = \frac{y}{2 x}$ (taking small angle approximation)

NEET- 2017

Ray Optics

282253 A light ray falls on a rectangular glass slab as shown. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is

1

\sqrt{3 / 2}

2

\frac{(\sqrt{3} + 1)}{2}

3

\frac{(\sqrt{2} + 1)}{2}

4

\sqrt{5} / 2

Explanation:

AGiven, $i = 45^{\circ}, r + θ_{c} = 90^{\circ}$ (from figure) $r =$ angle of refraction.
$θ_{c} =$ critical angle
Using Snell's law-
$μ = \frac{Sin i}{Sin r}$
$Sin r = \frac{Sin 45^{\circ}}{μ} = \frac{1}{\sqrt{2} μ}$
$Sin (90^{\circ} - θ_{c}) = \frac{1}{\sqrt{2} μ}$
$\frac{\sqrt{μ^{2} - 1}}{μ} = \frac{1}{\sqrt{2} μ}$
$\Rightarrow μ^{2} - 1 = \frac{1}{2}$
$\Rightarrow μ = \sqrt{3 / 2}$

AIIMS-27.05.2018(M)

Ray Optics

282254 An optical illusion which occurs mainly in deserts during hot summer is based on the principle of

1 Reflection

2 Interference

3 Dispersion

4 Total internal reflection

NDA (I) 2017

Ray Optics

282255 The critical angle for glass is $41^{\circ} 48^{'}$ and that for water is $48^{\circ} 36^{'}$ . Calculate the critical angle for glass-water interface.

1

62^{\circ} 43^{'}

2

34^{\circ} 42^{'}

3

52^{\circ} 42^{'}

4

44^{\circ} 42^{'}

Explanation:

A: Given that,
Critical angle for glass $(C_{g}) = 41^{\circ} 48^{'}$
Critical angle for water $(C_{w}) = 48^{\circ} 36^{'}$
Critical angle of glass water interface is
$\sin C_{g / w} = \frac{1}{μ_{g} / μ_{w}} = \frac{μ_{w}}{μ_{g}}$
For glass, $\sin C_{g} = \frac{1}{μ_{g}}$
For water $\sin C_{w} = \frac{1}{μ_{w}}$
$∴ \sin C_{g / w} = \frac{\sin C_{g}}{\sin C_{w}} = \frac{\sin {41.48}^{'}}{\sin {48.36}^{'}}$
Convert minute into degree. Minute is multiplied by $\frac{1}{60}$
$\begin{array}{r} = \frac{\sin {41.8}^{\circ}}{\sin {48.6}^{\circ}} = 0.89 \\ C_{g / w} = {62.87}^{\circ} ◻ 62^{\circ} 43^{'} \end{array}$

JCECE-2017

Ray Optics

282256 A beam of light from a source $L$ is incident normally on a plane mirror mixed at a certain distance $x$ from the source. The beam is reflected back as a spot on a scale placed just above the source $L$ . When the mirror is rotated through a small angle $θ$ , the spot of the light is found to move through a distance $y$ on the scale. The angle $θ$ is given by

1

\frac{y}{2 x}

2

\frac{y}{x}

3

\frac{x}{2 y}

4

\frac{x}{y}

Explanation:

A: According to question,

If mirror is rotated by angle $θ$ then angle of reflection will be rotated by $2 θ$ (because of law of reflection) Therefore, $\tan 2 θ = \frac{y}{x}$ $2 θ = \frac{y}{x}$ $θ = \frac{y}{2 x}$ (taking small angle approximation)

NEET- 2017

Ray Optics

282253 A light ray falls on a rectangular glass slab as shown. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is

1

\sqrt{3 / 2}

2

\frac{(\sqrt{3} + 1)}{2}

3

\frac{(\sqrt{2} + 1)}{2}

4

\sqrt{5} / 2

Explanation:

AGiven, $i = 45^{\circ}, r + θ_{c} = 90^{\circ}$ (from figure) $r =$ angle of refraction.
$θ_{c} =$ critical angle
Using Snell's law-
$μ = \frac{Sin i}{Sin r}$
$Sin r = \frac{Sin 45^{\circ}}{μ} = \frac{1}{\sqrt{2} μ}$
$Sin (90^{\circ} - θ_{c}) = \frac{1}{\sqrt{2} μ}$
$\frac{\sqrt{μ^{2} - 1}}{μ} = \frac{1}{\sqrt{2} μ}$
$\Rightarrow μ^{2} - 1 = \frac{1}{2}$
$\Rightarrow μ = \sqrt{3 / 2}$

AIIMS-27.05.2018(M)

Ray Optics

282254 An optical illusion which occurs mainly in deserts during hot summer is based on the principle of

1 Reflection

2 Interference

3 Dispersion

4 Total internal reflection

NDA (I) 2017

Ray Optics

282255 The critical angle for glass is $41^{\circ} 48^{'}$ and that for water is $48^{\circ} 36^{'}$ . Calculate the critical angle for glass-water interface.

1

62^{\circ} 43^{'}

2

34^{\circ} 42^{'}

3

52^{\circ} 42^{'}

4

44^{\circ} 42^{'}

Explanation:

A: Given that,
Critical angle for glass $(C_{g}) = 41^{\circ} 48^{'}$
Critical angle for water $(C_{w}) = 48^{\circ} 36^{'}$
Critical angle of glass water interface is
$\sin C_{g / w} = \frac{1}{μ_{g} / μ_{w}} = \frac{μ_{w}}{μ_{g}}$
For glass, $\sin C_{g} = \frac{1}{μ_{g}}$
For water $\sin C_{w} = \frac{1}{μ_{w}}$
$∴ \sin C_{g / w} = \frac{\sin C_{g}}{\sin C_{w}} = \frac{\sin {41.48}^{'}}{\sin {48.36}^{'}}$
Convert minute into degree. Minute is multiplied by $\frac{1}{60}$
$\begin{array}{r} = \frac{\sin {41.8}^{\circ}}{\sin {48.6}^{\circ}} = 0.89 \\ C_{g / w} = {62.87}^{\circ} ◻ 62^{\circ} 43^{'} \end{array}$

JCECE-2017

Ray Optics

282256 A beam of light from a source $L$ is incident normally on a plane mirror mixed at a certain distance $x$ from the source. The beam is reflected back as a spot on a scale placed just above the source $L$ . When the mirror is rotated through a small angle $θ$ , the spot of the light is found to move through a distance $y$ on the scale. The angle $θ$ is given by

1

\frac{y}{2 x}

2

\frac{y}{x}

3

\frac{x}{2 y}

4

\frac{x}{y}

Explanation:

A: According to question,

If mirror is rotated by angle $θ$ then angle of reflection will be rotated by $2 θ$ (because of law of reflection) Therefore, $\tan 2 θ = \frac{y}{x}$ $2 θ = \frac{y}{x}$ $θ = \frac{y}{2 x}$ (taking small angle approximation)

NEET- 2017

Ray Optics

282253 A light ray falls on a rectangular glass slab as shown. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is

1

\sqrt{3 / 2}

2

\frac{(\sqrt{3} + 1)}{2}

3

\frac{(\sqrt{2} + 1)}{2}

4

\sqrt{5} / 2

Explanation:

AGiven, $i = 45^{\circ}, r + θ_{c} = 90^{\circ}$ (from figure) $r =$ angle of refraction.
$θ_{c} =$ critical angle
Using Snell's law-
$μ = \frac{Sin i}{Sin r}$
$Sin r = \frac{Sin 45^{\circ}}{μ} = \frac{1}{\sqrt{2} μ}$
$Sin (90^{\circ} - θ_{c}) = \frac{1}{\sqrt{2} μ}$
$\frac{\sqrt{μ^{2} - 1}}{μ} = \frac{1}{\sqrt{2} μ}$
$\Rightarrow μ^{2} - 1 = \frac{1}{2}$
$\Rightarrow μ = \sqrt{3 / 2}$

AIIMS-27.05.2018(M)

Ray Optics

282254 An optical illusion which occurs mainly in deserts during hot summer is based on the principle of

1 Reflection

2 Interference

3 Dispersion

4 Total internal reflection

NDA (I) 2017

Ray Optics

282255 The critical angle for glass is $41^{\circ} 48^{'}$ and that for water is $48^{\circ} 36^{'}$ . Calculate the critical angle for glass-water interface.

1

62^{\circ} 43^{'}

2

34^{\circ} 42^{'}

3

52^{\circ} 42^{'}

4

44^{\circ} 42^{'}

Explanation:

A: Given that,
Critical angle for glass $(C_{g}) = 41^{\circ} 48^{'}$
Critical angle for water $(C_{w}) = 48^{\circ} 36^{'}$
Critical angle of glass water interface is
$\sin C_{g / w} = \frac{1}{μ_{g} / μ_{w}} = \frac{μ_{w}}{μ_{g}}$
For glass, $\sin C_{g} = \frac{1}{μ_{g}}$
For water $\sin C_{w} = \frac{1}{μ_{w}}$
$∴ \sin C_{g / w} = \frac{\sin C_{g}}{\sin C_{w}} = \frac{\sin {41.48}^{'}}{\sin {48.36}^{'}}$
Convert minute into degree. Minute is multiplied by $\frac{1}{60}$
$\begin{array}{r} = \frac{\sin {41.8}^{\circ}}{\sin {48.6}^{\circ}} = 0.89 \\ C_{g / w} = {62.87}^{\circ} ◻ 62^{\circ} 43^{'} \end{array}$

JCECE-2017

Ray Optics

282256 A beam of light from a source $L$ is incident normally on a plane mirror mixed at a certain distance $x$ from the source. The beam is reflected back as a spot on a scale placed just above the source $L$ . When the mirror is rotated through a small angle $θ$ , the spot of the light is found to move through a distance $y$ on the scale. The angle $θ$ is given by

1

\frac{y}{2 x}

2

\frac{y}{x}

3

\frac{x}{2 y}

4

\frac{x}{y}

Explanation:

A: According to question,

If mirror is rotated by angle $θ$ then angle of reflection will be rotated by $2 θ$ (because of law of reflection) Therefore, $\tan 2 θ = \frac{y}{x}$ $2 θ = \frac{y}{x}$ $θ = \frac{y}{2 x}$ (taking small angle approximation)

NEET- 2017