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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364765 Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$ . When the space between the two lenses is filled with glycerine (which has the same refractive index ( $μ = 1.5$ ) as that of glass) then the equivalent focal length is $F_{2}$ . The ratio $F_{1} : F_{2}$ will be :

1

2 : 1

2

1 : 2

3

2 : 3

4

3 : 4

Explanation:

supporting img
Equivalent focal length in air $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$
When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)
$\frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f} - \frac{1}{f} = \frac{1}{f}$
$\frac{F_{1}}{F_{2}} = \frac{1}{2}$

NEET - 2019

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364766 Two similar plano convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be
supporting img

1

1 : 1 : 1

2

2 : 2 : 1

3

2 : 1 : 1

4

1 : 2 : 2

Explanation:

Focal length of a plano convex lens of radius of curvature R and refractive index $μ$ is
For a plano convex lens.
supporting img
$\frac{1}{f} = (μ - 1) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ - 1}{R}$
Even for the convexo plane lens

$\frac{1}{f} = μ - 1 (\frac{1}{R} - \frac{1}{\infty}) = \frac{μ - 1}{R}$
For all the three given combinations
$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2 (μ - 1)}{R}$
$\Rightarrow F_{1} : F_{2} : F_{3} = 1 : 1 : 1$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364767 A layered lens as shown in figure is made of two types of transparent materials indicated by different shades. A point object is placed on its axis. The object will form
supporting img

1 3 image

2 9 image

3 1 image

4 2 image

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364768 An equiconvex lens of radius of curvature $14 c m$ is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of $40 c m$ , calculate the image distance.
supporting img

1

40 c m

2

25 c m

3

50 c m

4

35 c m

Explanation:

Effective focal length.
Focal length of any lens is given by
$\frac{1}{f} = (\frac{μ_{1} - μ_{m}}{μ_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Where $f =$ focal length of lens
$μ_{1} =$ Refractive index of less material
$μ_{m} =$ Refractive index of medium
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{(μ_{1} - 1)}{R} + \frac{(μ_{2} - 1)}{R}$
$\Rightarrow f = 20 c m$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{20} = \frac{1}{v} - (\frac{1}{- 40}) \Rightarrow v = 40 c m$

KCET - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364765 Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$ . When the space between the two lenses is filled with glycerine (which has the same refractive index ( $μ = 1.5$ ) as that of glass) then the equivalent focal length is $F_{2}$ . The ratio $F_{1} : F_{2}$ will be :

1

2 : 1

2

1 : 2

3

2 : 3

4

3 : 4

Explanation:

supporting img
Equivalent focal length in air $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$
When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)
$\frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f} - \frac{1}{f} = \frac{1}{f}$
$\frac{F_{1}}{F_{2}} = \frac{1}{2}$

NEET - 2019

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364766 Two similar plano convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be
supporting img

1

1 : 1 : 1

2

2 : 2 : 1

3

2 : 1 : 1

4

1 : 2 : 2

Explanation:

Focal length of a plano convex lens of radius of curvature R and refractive index $μ$ is
For a plano convex lens.
supporting img
$\frac{1}{f} = (μ - 1) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ - 1}{R}$
Even for the convexo plane lens

$\frac{1}{f} = μ - 1 (\frac{1}{R} - \frac{1}{\infty}) = \frac{μ - 1}{R}$
For all the three given combinations
$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2 (μ - 1)}{R}$
$\Rightarrow F_{1} : F_{2} : F_{3} = 1 : 1 : 1$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364767 A layered lens as shown in figure is made of two types of transparent materials indicated by different shades. A point object is placed on its axis. The object will form
supporting img

1 3 image

2 9 image

3 1 image

4 2 image

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364768 An equiconvex lens of radius of curvature $14 c m$ is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of $40 c m$ , calculate the image distance.
supporting img

1

40 c m

2

25 c m

3

50 c m

4

35 c m

Explanation:

Effective focal length.
Focal length of any lens is given by
$\frac{1}{f} = (\frac{μ_{1} - μ_{m}}{μ_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Where $f =$ focal length of lens
$μ_{1} =$ Refractive index of less material
$μ_{m} =$ Refractive index of medium
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{(μ_{1} - 1)}{R} + \frac{(μ_{2} - 1)}{R}$
$\Rightarrow f = 20 c m$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{20} = \frac{1}{v} - (\frac{1}{- 40}) \Rightarrow v = 40 c m$

KCET - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364765 Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$ . When the space between the two lenses is filled with glycerine (which has the same refractive index ( $μ = 1.5$ ) as that of glass) then the equivalent focal length is $F_{2}$ . The ratio $F_{1} : F_{2}$ will be :

1

2 : 1

2

1 : 2

3

2 : 3

4

3 : 4

Explanation:

supporting img
Equivalent focal length in air $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$
When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)
$\frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f} - \frac{1}{f} = \frac{1}{f}$
$\frac{F_{1}}{F_{2}} = \frac{1}{2}$

NEET - 2019

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364766 Two similar plano convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be
supporting img

1

1 : 1 : 1

2

2 : 2 : 1

3

2 : 1 : 1

4

1 : 2 : 2

Explanation:

Focal length of a plano convex lens of radius of curvature R and refractive index $μ$ is
For a plano convex lens.
supporting img
$\frac{1}{f} = (μ - 1) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ - 1}{R}$
Even for the convexo plane lens

$\frac{1}{f} = μ - 1 (\frac{1}{R} - \frac{1}{\infty}) = \frac{μ - 1}{R}$
For all the three given combinations
$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2 (μ - 1)}{R}$
$\Rightarrow F_{1} : F_{2} : F_{3} = 1 : 1 : 1$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364767 A layered lens as shown in figure is made of two types of transparent materials indicated by different shades. A point object is placed on its axis. The object will form
supporting img

1 3 image

2 9 image

3 1 image

4 2 image

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364768 An equiconvex lens of radius of curvature $14 c m$ is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of $40 c m$ , calculate the image distance.
supporting img

1

40 c m

2

25 c m

3

50 c m

4

35 c m

Explanation:

Effective focal length.
Focal length of any lens is given by
$\frac{1}{f} = (\frac{μ_{1} - μ_{m}}{μ_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Where $f =$ focal length of lens
$μ_{1} =$ Refractive index of less material
$μ_{m} =$ Refractive index of medium
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{(μ_{1} - 1)}{R} + \frac{(μ_{2} - 1)}{R}$
$\Rightarrow f = 20 c m$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{20} = \frac{1}{v} - (\frac{1}{- 40}) \Rightarrow v = 40 c m$

KCET - 2024

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364765 Two similar thin equi-convex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$ . When the space between the two lenses is filled with glycerine (which has the same refractive index ( $μ = 1.5$ ) as that of glass) then the equivalent focal length is $F_{2}$ . The ratio $F_{1} : F_{2}$ will be :

1

2 : 1

2

1 : 2

3

2 : 3

4

3 : 4

Explanation:

supporting img
Equivalent focal length in air $\frac{1}{F_{1}} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$
When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)
$\frac{1}{F_{2}} = \frac{1}{f} + \frac{1}{f} - \frac{1}{f} = \frac{1}{f}$
$\frac{F_{1}}{F_{2}} = \frac{1}{2}$

NEET - 2019

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364766 Two similar plano convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be
supporting img

1

1 : 1 : 1

2

2 : 2 : 1

3

2 : 1 : 1

4

1 : 2 : 2

Explanation:

Focal length of a plano convex lens of radius of curvature R and refractive index $μ$ is
For a plano convex lens.
supporting img
$\frac{1}{f} = (μ - 1) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ - 1}{R}$
Even for the convexo plane lens

$\frac{1}{f} = μ - 1 (\frac{1}{R} - \frac{1}{\infty}) = \frac{μ - 1}{R}$
For all the three given combinations
$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2 (μ - 1)}{R}$
$\Rightarrow F_{1} : F_{2} : F_{3} = 1 : 1 : 1$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364767 A layered lens as shown in figure is made of two types of transparent materials indicated by different shades. A point object is placed on its axis. The object will form
supporting img

1 3 image

2 9 image

3 1 image

4 2 image

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364768 An equiconvex lens of radius of curvature $14 c m$ is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of $40 c m$ , calculate the image distance.
supporting img

1

40 c m

2

25 c m

3

50 c m

4

35 c m

Explanation:

Effective focal length.
Focal length of any lens is given by
$\frac{1}{f} = (\frac{μ_{1} - μ_{m}}{μ_{m}}) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Where $f =$ focal length of lens
$μ_{1} =$ Refractive index of less material
$μ_{m} =$ Refractive index of medium
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{(μ_{1} - 1)}{R} + \frac{(μ_{2} - 1)}{R}$
$\Rightarrow f = 20 c m$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{20} = \frac{1}{v} - (\frac{1}{- 40}) \Rightarrow v = 40 c m$

KCET - 2024

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