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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364795 A bioconvex lens has radii of curvature, 20 $c m$ each. If the refractive index of the material of the lens is 1.5, the power of the lens is

1 +20 D

2 +5D

3 infinity

4 +2D

Explanation:

$R_{1} = R_{2} = 20 c m = 0.2 m$
$μ = \frac{3}{2}$
$P = \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Take sign as +ve in incident ray direction
supporting img
$P = (\frac{3}{2} - 1) (\frac{1}{0.2} + \frac{1}{0.2})$
$P = \frac{1}{2} (\frac{2}{0.2}) = \frac{10}{2} = + 5 D$

NEET - 2022

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364796 The power of a thin convex lens $(_{a} n_{g} = 1.5)$ is $5.0 D .$ When it is placed in a liquid of refractive index $_{a} n_{l}$ , then it behaves as a concave lens of focal length $100 c m$ . The refractive index of the liquid $_{a} n_{l}$ will be

1

5 / 3

2

4 / 3

3

\sqrt{3}

4

5 / 4

Explanation:

According to lens maker formula,
$\frac{1}{f} = (\frac{n_{g} - n_{m}}{n_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (1)$
$H e r e,_{a} n_{g} = \frac{n_{g}}{n_{a}} = 1.5$
$P = \frac{1}{f} = 5 D$
$∴ 5 = (1.5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (2)$
Also, $f = - 100 c m = - 1 m, n_{l} = ?$
$∴ \frac{1}{- 1} = (\frac{1.5}{n_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (3)$
From eqn. (2) and (3),
$- 5 = \frac{0.5}{(\frac{1.5}{n_{l}} - 1)} \Rightarrow n_{l} = \frac{5}{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364797 A narrow parallel beam of light is incident paraxially on a solid transparent sphere of radius $r$ kept in air. What should be the refractive index if the beam is to be focused at the farther surface of the sphere.

1 1.5

2 2

3 1.3

4 None

Explanation:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
For refraction at first surface
$μ_{1} = 1 μ_{2} = μ$
$u = \infty R = R$
$v = 2 R$
$\frac{μ}{2 R} - \frac{1}{\infty} = \frac{μ - 1}{R}$
$\frac{μ}{2} = μ - 1 \Rightarrow μ = 1.5$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364798 An air bubble is contained inside water. It behaves as a
supporting img

1 Convex lens

2 Concave lens

3 Cannot say

4 Neither convex nor concave

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364795 A bioconvex lens has radii of curvature, 20 $c m$ each. If the refractive index of the material of the lens is 1.5, the power of the lens is

1 +20 D

2 +5D

3 infinity

4 +2D

Explanation:

$R_{1} = R_{2} = 20 c m = 0.2 m$
$μ = \frac{3}{2}$
$P = \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Take sign as +ve in incident ray direction
supporting img
$P = (\frac{3}{2} - 1) (\frac{1}{0.2} + \frac{1}{0.2})$
$P = \frac{1}{2} (\frac{2}{0.2}) = \frac{10}{2} = + 5 D$

NEET - 2022

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364796 The power of a thin convex lens $(_{a} n_{g} = 1.5)$ is $5.0 D .$ When it is placed in a liquid of refractive index $_{a} n_{l}$ , then it behaves as a concave lens of focal length $100 c m$ . The refractive index of the liquid $_{a} n_{l}$ will be

1

5 / 3

2

4 / 3

3

\sqrt{3}

4

5 / 4

Explanation:

According to lens maker formula,
$\frac{1}{f} = (\frac{n_{g} - n_{m}}{n_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (1)$
$H e r e,_{a} n_{g} = \frac{n_{g}}{n_{a}} = 1.5$
$P = \frac{1}{f} = 5 D$
$∴ 5 = (1.5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (2)$
Also, $f = - 100 c m = - 1 m, n_{l} = ?$
$∴ \frac{1}{- 1} = (\frac{1.5}{n_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (3)$
From eqn. (2) and (3),
$- 5 = \frac{0.5}{(\frac{1.5}{n_{l}} - 1)} \Rightarrow n_{l} = \frac{5}{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364797 A narrow parallel beam of light is incident paraxially on a solid transparent sphere of radius $r$ kept in air. What should be the refractive index if the beam is to be focused at the farther surface of the sphere.

1 1.5

2 2

3 1.3

4 None

Explanation:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
For refraction at first surface
$μ_{1} = 1 μ_{2} = μ$
$u = \infty R = R$
$v = 2 R$
$\frac{μ}{2 R} - \frac{1}{\infty} = \frac{μ - 1}{R}$
$\frac{μ}{2} = μ - 1 \Rightarrow μ = 1.5$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364798 An air bubble is contained inside water. It behaves as a
supporting img

1 Convex lens

2 Concave lens

3 Cannot say

4 Neither convex nor concave

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364795 A bioconvex lens has radii of curvature, 20 $c m$ each. If the refractive index of the material of the lens is 1.5, the power of the lens is

1 +20 D

2 +5D

3 infinity

4 +2D

Explanation:

$R_{1} = R_{2} = 20 c m = 0.2 m$
$μ = \frac{3}{2}$
$P = \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Take sign as +ve in incident ray direction
supporting img
$P = (\frac{3}{2} - 1) (\frac{1}{0.2} + \frac{1}{0.2})$
$P = \frac{1}{2} (\frac{2}{0.2}) = \frac{10}{2} = + 5 D$

NEET - 2022

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364796 The power of a thin convex lens $(_{a} n_{g} = 1.5)$ is $5.0 D .$ When it is placed in a liquid of refractive index $_{a} n_{l}$ , then it behaves as a concave lens of focal length $100 c m$ . The refractive index of the liquid $_{a} n_{l}$ will be

1

5 / 3

2

4 / 3

3

\sqrt{3}

4

5 / 4

Explanation:

According to lens maker formula,
$\frac{1}{f} = (\frac{n_{g} - n_{m}}{n_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (1)$
$H e r e,_{a} n_{g} = \frac{n_{g}}{n_{a}} = 1.5$
$P = \frac{1}{f} = 5 D$
$∴ 5 = (1.5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (2)$
Also, $f = - 100 c m = - 1 m, n_{l} = ?$
$∴ \frac{1}{- 1} = (\frac{1.5}{n_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (3)$
From eqn. (2) and (3),
$- 5 = \frac{0.5}{(\frac{1.5}{n_{l}} - 1)} \Rightarrow n_{l} = \frac{5}{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364797 A narrow parallel beam of light is incident paraxially on a solid transparent sphere of radius $r$ kept in air. What should be the refractive index if the beam is to be focused at the farther surface of the sphere.

1 1.5

2 2

3 1.3

4 None

Explanation:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
For refraction at first surface
$μ_{1} = 1 μ_{2} = μ$
$u = \infty R = R$
$v = 2 R$
$\frac{μ}{2 R} - \frac{1}{\infty} = \frac{μ - 1}{R}$
$\frac{μ}{2} = μ - 1 \Rightarrow μ = 1.5$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364798 An air bubble is contained inside water. It behaves as a
supporting img

1 Convex lens

2 Concave lens

3 Cannot say

4 Neither convex nor concave

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364795 A bioconvex lens has radii of curvature, 20 $c m$ each. If the refractive index of the material of the lens is 1.5, the power of the lens is

1 +20 D

2 +5D

3 infinity

4 +2D

Explanation:

$R_{1} = R_{2} = 20 c m = 0.2 m$
$μ = \frac{3}{2}$
$P = \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Take sign as +ve in incident ray direction
supporting img
$P = (\frac{3}{2} - 1) (\frac{1}{0.2} + \frac{1}{0.2})$
$P = \frac{1}{2} (\frac{2}{0.2}) = \frac{10}{2} = + 5 D$

NEET - 2022

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364796 The power of a thin convex lens $(_{a} n_{g} = 1.5)$ is $5.0 D .$ When it is placed in a liquid of refractive index $_{a} n_{l}$ , then it behaves as a concave lens of focal length $100 c m$ . The refractive index of the liquid $_{a} n_{l}$ will be

1

5 / 3

2

4 / 3

3

\sqrt{3}

4

5 / 4

Explanation:

According to lens maker formula,
$\frac{1}{f} = (\frac{n_{g} - n_{m}}{n_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (1)$
$H e r e,_{a} n_{g} = \frac{n_{g}}{n_{a}} = 1.5$
$P = \frac{1}{f} = 5 D$
$∴ 5 = (1.5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (2)$
Also, $f = - 100 c m = - 1 m, n_{l} = ?$
$∴ \frac{1}{- 1} = (\frac{1.5}{n_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) (3)$
From eqn. (2) and (3),
$- 5 = \frac{0.5}{(\frac{1.5}{n_{l}} - 1)} \Rightarrow n_{l} = \frac{5}{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364797 A narrow parallel beam of light is incident paraxially on a solid transparent sphere of radius $r$ kept in air. What should be the refractive index if the beam is to be focused at the farther surface of the sphere.

1 1.5

2 2

3 1.3

4 None

Explanation:

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
For refraction at first surface
$μ_{1} = 1 μ_{2} = μ$
$u = \infty R = R$
$v = 2 R$
$\frac{μ}{2 R} - \frac{1}{\infty} = \frac{μ - 1}{R}$
$\frac{μ}{2} = μ - 1 \Rightarrow μ = 1.5$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364798 An air bubble is contained inside water. It behaves as a
supporting img

1 Convex lens

2 Concave lens

3 Cannot say

4 Neither convex nor concave

Explanation:

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