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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364855 A convex and concave lens separated by distance $d$ are then put in contact. The focal length of the combination

1 Becomes zero

2 Remains the same

3 Decreases

4 Increases

Explanation:

Focal length of a combination with separation (d) can be calculated as, $\frac{1}{F_{1}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} (1)$
Focal length of a combination without separation can be calculated as,
$\frac{1}{F_{2}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} (2)$
From (1) and (2),
$\frac{1}{F_{1}} = \frac{1}{F_{2}} - \frac{d}{f_{1} f_{2}} \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} = \frac{d}{f_{1} f_{2}}$
Since $f_{1}$ and $f_{2}$ have opposite signs,
$\frac{d}{f_{1} f_{2}} < 0 \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} < 0 \Rightarrow F_{2} > F_{1}$

KCET - 2008

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364856 The radii of curvature of the two surfaces of a lens are $20 c m$ and $30 c m$ and the refractive index of the material of the lens is 1.5 . If the lens is concavo-convex, then the focal length of the lens is:

1

24 c m

2

10 c m

3

15 c m

4

120 c m

Explanation:

The focal length of the lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f} = (1.5 - 1) (\frac{1}{20} - \frac{1}{30})$
$\Rightarrow \frac{1}{f} = 0.5 (\frac{30 - 20}{600})$
$\Rightarrow f = 120 c m$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364857 The radius of curvature for a convex lens is $40 c m$ for each surface. Its refractive index is 1.5. The focal length will be:

1

40 c m

2

20 c m

3

80 c m

4

30 c m

Explanation:

By according to the lens maker formula
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\frac{1}{f} = \frac{1}{2} [\frac{2}{40}] \Rightarrow f = 40 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364858 A 2 meter long scale with least count of $0.2 c m$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 c m$ mark and $1 m$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 c m$ mark. The $%$ error in the estimation of focal length is

1 1.70

2 1.02

3 0.51

4 0.85

Explanation:

$A s, u = (100 \pm 0.2) c m - (80 \pm 0.2) c m$
$= (20 \pm 0.4) c m$
$v = (180 \pm 0.2) - (100 \pm 0.2) = (80 \pm 0.4) c m$ .
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{5}{80}$
$f = 16 c m$
Using sign convention for real image, real object and convex lens, $\frac{1}{(+ v)} - \frac{1}{(- u)} = \frac{1}{(+ f)}$
Differentiating,
$\frac{1}{v^{2}} Δ v + \frac{1}{u^{2}} Δ u = \frac{1}{f^{2}} Δ f$
$\frac{Δ f}{f^{2}} = \frac{0.4}{6400} + \frac{0.4}{400}$
$\frac{Δ f}{f} = [\frac{2720}{6400 \times 400}] f$
$\frac{Δ f}{f} \times 100 = 0.017 \times 100 = 1.70 %$
supporting img

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364855 A convex and concave lens separated by distance $d$ are then put in contact. The focal length of the combination

1 Becomes zero

2 Remains the same

3 Decreases

4 Increases

Explanation:

Focal length of a combination with separation (d) can be calculated as, $\frac{1}{F_{1}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} (1)$
Focal length of a combination without separation can be calculated as,
$\frac{1}{F_{2}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} (2)$
From (1) and (2),
$\frac{1}{F_{1}} = \frac{1}{F_{2}} - \frac{d}{f_{1} f_{2}} \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} = \frac{d}{f_{1} f_{2}}$
Since $f_{1}$ and $f_{2}$ have opposite signs,
$\frac{d}{f_{1} f_{2}} < 0 \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} < 0 \Rightarrow F_{2} > F_{1}$

KCET - 2008

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364856 The radii of curvature of the two surfaces of a lens are $20 c m$ and $30 c m$ and the refractive index of the material of the lens is 1.5 . If the lens is concavo-convex, then the focal length of the lens is:

1

24 c m

2

10 c m

3

15 c m

4

120 c m

Explanation:

The focal length of the lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f} = (1.5 - 1) (\frac{1}{20} - \frac{1}{30})$
$\Rightarrow \frac{1}{f} = 0.5 (\frac{30 - 20}{600})$
$\Rightarrow f = 120 c m$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364857 The radius of curvature for a convex lens is $40 c m$ for each surface. Its refractive index is 1.5. The focal length will be:

1

40 c m

2

20 c m

3

80 c m

4

30 c m

Explanation:

By according to the lens maker formula
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\frac{1}{f} = \frac{1}{2} [\frac{2}{40}] \Rightarrow f = 40 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364858 A 2 meter long scale with least count of $0.2 c m$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 c m$ mark and $1 m$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 c m$ mark. The $%$ error in the estimation of focal length is

1 1.70

2 1.02

3 0.51

4 0.85

Explanation:

$A s, u = (100 \pm 0.2) c m - (80 \pm 0.2) c m$
$= (20 \pm 0.4) c m$
$v = (180 \pm 0.2) - (100 \pm 0.2) = (80 \pm 0.4) c m$ .
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{5}{80}$
$f = 16 c m$
Using sign convention for real image, real object and convex lens, $\frac{1}{(+ v)} - \frac{1}{(- u)} = \frac{1}{(+ f)}$
Differentiating,
$\frac{1}{v^{2}} Δ v + \frac{1}{u^{2}} Δ u = \frac{1}{f^{2}} Δ f$
$\frac{Δ f}{f^{2}} = \frac{0.4}{6400} + \frac{0.4}{400}$
$\frac{Δ f}{f} = [\frac{2720}{6400 \times 400}] f$
$\frac{Δ f}{f} \times 100 = 0.017 \times 100 = 1.70 %$
supporting img

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364855 A convex and concave lens separated by distance $d$ are then put in contact. The focal length of the combination

1 Becomes zero

2 Remains the same

3 Decreases

4 Increases

Explanation:

Focal length of a combination with separation (d) can be calculated as, $\frac{1}{F_{1}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} (1)$
Focal length of a combination without separation can be calculated as,
$\frac{1}{F_{2}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} (2)$
From (1) and (2),
$\frac{1}{F_{1}} = \frac{1}{F_{2}} - \frac{d}{f_{1} f_{2}} \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} = \frac{d}{f_{1} f_{2}}$
Since $f_{1}$ and $f_{2}$ have opposite signs,
$\frac{d}{f_{1} f_{2}} < 0 \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} < 0 \Rightarrow F_{2} > F_{1}$

KCET - 2008

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364856 The radii of curvature of the two surfaces of a lens are $20 c m$ and $30 c m$ and the refractive index of the material of the lens is 1.5 . If the lens is concavo-convex, then the focal length of the lens is:

1

24 c m

2

10 c m

3

15 c m

4

120 c m

Explanation:

The focal length of the lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f} = (1.5 - 1) (\frac{1}{20} - \frac{1}{30})$
$\Rightarrow \frac{1}{f} = 0.5 (\frac{30 - 20}{600})$
$\Rightarrow f = 120 c m$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364857 The radius of curvature for a convex lens is $40 c m$ for each surface. Its refractive index is 1.5. The focal length will be:

1

40 c m

2

20 c m

3

80 c m

4

30 c m

Explanation:

By according to the lens maker formula
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\frac{1}{f} = \frac{1}{2} [\frac{2}{40}] \Rightarrow f = 40 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364858 A 2 meter long scale with least count of $0.2 c m$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 c m$ mark and $1 m$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 c m$ mark. The $%$ error in the estimation of focal length is

1 1.70

2 1.02

3 0.51

4 0.85

Explanation:

$A s, u = (100 \pm 0.2) c m - (80 \pm 0.2) c m$
$= (20 \pm 0.4) c m$
$v = (180 \pm 0.2) - (100 \pm 0.2) = (80 \pm 0.4) c m$ .
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{5}{80}$
$f = 16 c m$
Using sign convention for real image, real object and convex lens, $\frac{1}{(+ v)} - \frac{1}{(- u)} = \frac{1}{(+ f)}$
Differentiating,
$\frac{1}{v^{2}} Δ v + \frac{1}{u^{2}} Δ u = \frac{1}{f^{2}} Δ f$
$\frac{Δ f}{f^{2}} = \frac{0.4}{6400} + \frac{0.4}{400}$
$\frac{Δ f}{f} = [\frac{2720}{6400 \times 400}] f$
$\frac{Δ f}{f} \times 100 = 0.017 \times 100 = 1.70 %$
supporting img

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364855 A convex and concave lens separated by distance $d$ are then put in contact. The focal length of the combination

1 Becomes zero

2 Remains the same

3 Decreases

4 Increases

Explanation:

Focal length of a combination with separation (d) can be calculated as, $\frac{1}{F_{1}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} (1)$
Focal length of a combination without separation can be calculated as,
$\frac{1}{F_{2}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} (2)$
From (1) and (2),
$\frac{1}{F_{1}} = \frac{1}{F_{2}} - \frac{d}{f_{1} f_{2}} \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} = \frac{d}{f_{1} f_{2}}$
Since $f_{1}$ and $f_{2}$ have opposite signs,
$\frac{d}{f_{1} f_{2}} < 0 \Rightarrow \frac{1}{F_{2}} - \frac{1}{F_{1}} < 0 \Rightarrow F_{2} > F_{1}$

KCET - 2008

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364856 The radii of curvature of the two surfaces of a lens are $20 c m$ and $30 c m$ and the refractive index of the material of the lens is 1.5 . If the lens is concavo-convex, then the focal length of the lens is:

1

24 c m

2

10 c m

3

15 c m

4

120 c m

Explanation:

The focal length of the lens,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\Rightarrow \frac{1}{f} = (1.5 - 1) (\frac{1}{20} - \frac{1}{30})$
$\Rightarrow \frac{1}{f} = 0.5 (\frac{30 - 20}{600})$
$\Rightarrow f = 120 c m$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364857 The radius of curvature for a convex lens is $40 c m$ for each surface. Its refractive index is 1.5. The focal length will be:

1

40 c m

2

20 c m

3

80 c m

4

30 c m

Explanation:

By according to the lens maker formula
$\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
$\frac{1}{f} = \frac{1}{2} [\frac{2}{40}] \Rightarrow f = 40 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364858 A 2 meter long scale with least count of $0.2 c m$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 c m$ mark and $1 m$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 c m$ mark. The $%$ error in the estimation of focal length is

1 1.70

2 1.02

3 0.51

4 0.85

Explanation:

$A s, u = (100 \pm 0.2) c m - (80 \pm 0.2) c m$
$= (20 \pm 0.4) c m$
$v = (180 \pm 0.2) - (100 \pm 0.2) = (80 \pm 0.4) c m$ .
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{5}{80}$
$f = 16 c m$
Using sign convention for real image, real object and convex lens, $\frac{1}{(+ v)} - \frac{1}{(- u)} = \frac{1}{(+ f)}$
Differentiating,
$\frac{1}{v^{2}} Δ v + \frac{1}{u^{2}} Δ u = \frac{1}{f^{2}} Δ f$
$\frac{Δ f}{f^{2}} = \frac{0.4}{6400} + \frac{0.4}{400}$
$\frac{Δ f}{f} = [\frac{2720}{6400 \times 400}] f$
$\frac{Δ f}{f} \times 100 = 0.017 \times 100 = 1.70 %$
supporting img

JEE - 2023

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