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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364829 When an object is at a distance $x$ and $y$ from the optical centre of a lens, a real and a virtual image are formed, respectively with same magnification. The focal length of lens is

1

x + y

2

(x + y) / 2

3

x + y / 2

4

\sqrt{x y}

Explanation:

We have, $m = \frac{f}{u - f}$
Let $m$ and $m^{'}$ be the magnification when object is at $x$ and $y$ respectively.
Then, $m = \frac{f}{- x - f}$ and $m^{'} = \frac{f}{- y - f}$
As $m^{'} = - m$
$\Rightarrow \frac{f}{- y - f} = \frac{- f}{- x - f}$
$\Rightarrow f = - (\frac{x + y}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364830 A convex lens made of glass has focal length $0.15 m$ in air. If the refractive index of glass is 3/2 and that of water is 4/3, the focal length of lens when immersed in water is

1

0.45 m

2

0.15 m

3

0.30 m

4

0.6 m

Explanation:

Here, $f_{a} = 0.15 m, μ_{g} = 3 / 2, μ_{w} = 4 / 3$
Using the lensmaker’s formula, when the lens is in air,
$\frac{1}{f_{a}} = (\frac{μ_{g}}{μ_{a}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{1} - 1) C$
where $C = (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f_{a}} = \frac{C}{2} (1)$
Also, when the lens is in water,
$\frac{1}{f_{w}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{(4 / 3)} - 1) C$
$\frac{1}{f_{w}} = \frac{C}{8} (2)$
From (1) and (2), we get
$\frac{f_{w}}{f_{a}} = \frac{C}{2} \times \frac{8}{C} = 4$
$f_{w} = 4 f_{a} = 4 \times 0.15 = 0.6 m$

KCET - 2009

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364831 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
Statement B :
Concave lens always forms a virtual and erect image.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct

4 Both Statements are incorrect.

Explanation:

If $u = - R = - 2 f$ then $v = - 2 f = - R$ .
The image formed by the concave lens is always on the same side of the lens. Concave lens always forms virtual and erect image. So the correct option is (2).

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364832 The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then, the refractive index of the material of the lens is

1 greater than zero but less than 1.5

2 greater than 1.5 but less than 2.0

3 greater than 2.0 but less than 2.5

4 greater than 2.5 but less than 2.0

Explanation:

Focal length of lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
For equi-convex
$R_{1} = + R, R_{2} = - R$
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (μ - 1) (\frac{2}{R})$
$f = \frac{R}{2 (μ - 1)} \Rightarrow f > R, so 2 (μ - 1) < 1$
$i . e . (μ - 1) < \frac{1}{2} \Rightarrow (μ - 1) < 0.5$
$\Rightarrow μ < 1.5$
Focal length of convex lens is positive,
so $u$ cannot be negative. Hence, $μ$
should be greater than zero but less
than 1.5.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364829 When an object is at a distance $x$ and $y$ from the optical centre of a lens, a real and a virtual image are formed, respectively with same magnification. The focal length of lens is

1

x + y

2

(x + y) / 2

3

x + y / 2

4

\sqrt{x y}

Explanation:

We have, $m = \frac{f}{u - f}$
Let $m$ and $m^{'}$ be the magnification when object is at $x$ and $y$ respectively.
Then, $m = \frac{f}{- x - f}$ and $m^{'} = \frac{f}{- y - f}$
As $m^{'} = - m$
$\Rightarrow \frac{f}{- y - f} = \frac{- f}{- x - f}$
$\Rightarrow f = - (\frac{x + y}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364830 A convex lens made of glass has focal length $0.15 m$ in air. If the refractive index of glass is 3/2 and that of water is 4/3, the focal length of lens when immersed in water is

1

0.45 m

2

0.15 m

3

0.30 m

4

0.6 m

Explanation:

Here, $f_{a} = 0.15 m, μ_{g} = 3 / 2, μ_{w} = 4 / 3$
Using the lensmaker’s formula, when the lens is in air,
$\frac{1}{f_{a}} = (\frac{μ_{g}}{μ_{a}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{1} - 1) C$
where $C = (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f_{a}} = \frac{C}{2} (1)$
Also, when the lens is in water,
$\frac{1}{f_{w}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{(4 / 3)} - 1) C$
$\frac{1}{f_{w}} = \frac{C}{8} (2)$
From (1) and (2), we get
$\frac{f_{w}}{f_{a}} = \frac{C}{2} \times \frac{8}{C} = 4$
$f_{w} = 4 f_{a} = 4 \times 0.15 = 0.6 m$

KCET - 2009

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364831 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
Statement B :
Concave lens always forms a virtual and erect image.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct

4 Both Statements are incorrect.

Explanation:

If $u = - R = - 2 f$ then $v = - 2 f = - R$ .
The image formed by the concave lens is always on the same side of the lens. Concave lens always forms virtual and erect image. So the correct option is (2).

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364832 The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then, the refractive index of the material of the lens is

1 greater than zero but less than 1.5

2 greater than 1.5 but less than 2.0

3 greater than 2.0 but less than 2.5

4 greater than 2.5 but less than 2.0

Explanation:

Focal length of lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
For equi-convex
$R_{1} = + R, R_{2} = - R$
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (μ - 1) (\frac{2}{R})$
$f = \frac{R}{2 (μ - 1)} \Rightarrow f > R, so 2 (μ - 1) < 1$
$i . e . (μ - 1) < \frac{1}{2} \Rightarrow (μ - 1) < 0.5$
$\Rightarrow μ < 1.5$
Focal length of convex lens is positive,
so $u$ cannot be negative. Hence, $μ$
should be greater than zero but less
than 1.5.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364829 When an object is at a distance $x$ and $y$ from the optical centre of a lens, a real and a virtual image are formed, respectively with same magnification. The focal length of lens is

1

x + y

2

(x + y) / 2

3

x + y / 2

4

\sqrt{x y}

Explanation:

We have, $m = \frac{f}{u - f}$
Let $m$ and $m^{'}$ be the magnification when object is at $x$ and $y$ respectively.
Then, $m = \frac{f}{- x - f}$ and $m^{'} = \frac{f}{- y - f}$
As $m^{'} = - m$
$\Rightarrow \frac{f}{- y - f} = \frac{- f}{- x - f}$
$\Rightarrow f = - (\frac{x + y}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364830 A convex lens made of glass has focal length $0.15 m$ in air. If the refractive index of glass is 3/2 and that of water is 4/3, the focal length of lens when immersed in water is

1

0.45 m

2

0.15 m

3

0.30 m

4

0.6 m

Explanation:

Here, $f_{a} = 0.15 m, μ_{g} = 3 / 2, μ_{w} = 4 / 3$
Using the lensmaker’s formula, when the lens is in air,
$\frac{1}{f_{a}} = (\frac{μ_{g}}{μ_{a}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{1} - 1) C$
where $C = (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f_{a}} = \frac{C}{2} (1)$
Also, when the lens is in water,
$\frac{1}{f_{w}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{(4 / 3)} - 1) C$
$\frac{1}{f_{w}} = \frac{C}{8} (2)$
From (1) and (2), we get
$\frac{f_{w}}{f_{a}} = \frac{C}{2} \times \frac{8}{C} = 4$
$f_{w} = 4 f_{a} = 4 \times 0.15 = 0.6 m$

KCET - 2009

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364831 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
Statement B :
Concave lens always forms a virtual and erect image.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct

4 Both Statements are incorrect.

Explanation:

If $u = - R = - 2 f$ then $v = - 2 f = - R$ .
The image formed by the concave lens is always on the same side of the lens. Concave lens always forms virtual and erect image. So the correct option is (2).

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364832 The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then, the refractive index of the material of the lens is

1 greater than zero but less than 1.5

2 greater than 1.5 but less than 2.0

3 greater than 2.0 but less than 2.5

4 greater than 2.5 but less than 2.0

Explanation:

Focal length of lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
For equi-convex
$R_{1} = + R, R_{2} = - R$
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (μ - 1) (\frac{2}{R})$
$f = \frac{R}{2 (μ - 1)} \Rightarrow f > R, so 2 (μ - 1) < 1$
$i . e . (μ - 1) < \frac{1}{2} \Rightarrow (μ - 1) < 0.5$
$\Rightarrow μ < 1.5$
Focal length of convex lens is positive,
so $u$ cannot be negative. Hence, $μ$
should be greater than zero but less
than 1.5.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364829 When an object is at a distance $x$ and $y$ from the optical centre of a lens, a real and a virtual image are formed, respectively with same magnification. The focal length of lens is

1

x + y

2

(x + y) / 2

3

x + y / 2

4

\sqrt{x y}

Explanation:

We have, $m = \frac{f}{u - f}$
Let $m$ and $m^{'}$ be the magnification when object is at $x$ and $y$ respectively.
Then, $m = \frac{f}{- x - f}$ and $m^{'} = \frac{f}{- y - f}$
As $m^{'} = - m$
$\Rightarrow \frac{f}{- y - f} = \frac{- f}{- x - f}$
$\Rightarrow f = - (\frac{x + y}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364830 A convex lens made of glass has focal length $0.15 m$ in air. If the refractive index of glass is 3/2 and that of water is 4/3, the focal length of lens when immersed in water is

1

0.45 m

2

0.15 m

3

0.30 m

4

0.6 m

Explanation:

Here, $f_{a} = 0.15 m, μ_{g} = 3 / 2, μ_{w} = 4 / 3$
Using the lensmaker’s formula, when the lens is in air,
$\frac{1}{f_{a}} = (\frac{μ_{g}}{μ_{a}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{1} - 1) C$
where $C = (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f_{a}} = \frac{C}{2} (1)$
Also, when the lens is in water,
$\frac{1}{f_{w}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (\frac{(3 / 2)}{(4 / 3)} - 1) C$
$\frac{1}{f_{w}} = \frac{C}{8} (2)$
From (1) and (2), we get
$\frac{f_{w}}{f_{a}} = \frac{C}{2} \times \frac{8}{C} = 4$
$f_{w} = 4 f_{a} = 4 \times 0.15 = 0.6 m$

KCET - 2009

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364831 Read the Statement -A and Statement -B carefully to mark the correct options given below:
Statement A :
When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
Statement B :
Concave lens always forms a virtual and erect image.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct

4 Both Statements are incorrect.

Explanation:

If $u = - R = - 2 f$ then $v = - 2 f = - R$ .
The image formed by the concave lens is always on the same side of the lens. Concave lens always forms virtual and erect image. So the correct option is (2).

JEE - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364832 The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then, the refractive index of the material of the lens is

1 greater than zero but less than 1.5

2 greater than 1.5 but less than 2.0

3 greater than 2.0 but less than 2.5

4 greater than 2.5 but less than 2.0

Explanation:

Focal length of lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
For equi-convex
$R_{1} = + R, R_{2} = - R$
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (μ - 1) (\frac{2}{R})$
$f = \frac{R}{2 (μ - 1)} \Rightarrow f > R, so 2 (μ - 1) < 1$
$i . e . (μ - 1) < \frac{1}{2} \Rightarrow (μ - 1) < 0.5$
$\Rightarrow μ < 1.5$
Focal length of convex lens is positive,
so $u$ cannot be negative. Hence, $μ$
should be greater than zero but less
than 1.5.

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