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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364778 Find the position of final image from first lens. Given, focal length of each lens is $10 c m .$
supporting img

1

40 c m

2

50 c m

3

45 c m

4

55 c m

Explanation:

supporting img
(1) For refraction from 1st lens,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{40} = \frac{3}{40} c m$
$\Rightarrow v = \frac{40}{3} c m$
This image acts as object for 2nd lens.
$∴$ Object distance for 2 nd lens $L_{2}$
$= 30 - \frac{40}{3} = \frac{50}{3} c m$
(2) Now for refraction from $L_{2}$ ,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{10} - \frac{3}{50} = \frac{2}{50}$
$\Rightarrow v = 25 c m$
This image is object for 3rd lens $L_{3}$
$(u = 30 - 25 = 5 c m)$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{5}$
$v = - 10 c m$
(location of final image $= 10 c m$ to
left of $L_{3}$ )
$∴$ Distance from $L_{1}$
$= 30 + (30 - 10) = 50 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364779 The focal length of a convex lens of refractive index 1.5 is $f$ . If it is cut into four equal parts as shown in the figure. The focal length of the one of the parts (Shaded) will be
supporting img

1

f

2

2 f

3

3 f

4

4 f

Explanation:

As $\frac{1}{f^{'}} = (μ - 1) (\frac{1}{R} - \frac{1}{\infty})$
$\frac{1}{f^{'}} = \frac{μ - 1}{R} (1)$
and, $\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R})$
$\frac{1}{f} = \frac{2 (μ - 1)}{R} (2)$
Dividing (2) by (1), we get $f^{'} = 2 f$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364780 A lens is cut from the optical centre along the principal axis and the two parts of the lens are placed sideways as shown in figure. Then the new focal length of the combination:
supporting img

1 Remains the same

2 Becomes double

3 Becomes half

4 Becomes triple

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364781 Assertion :
A bi-convex lens of focal length $10 c m$ is split into two equals parts by a plane parallel to its principal axis. The focal length of the each part will be $20 c m$
Reason :
Focal length does not depend on the radii of curvature of two surfaces.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

A bi-convex lens with a focal length of $10 c m$ is split into two equal parts by a plane parallel to its principal axis, and the focal length of each part will be $20 c m$ . This is incorrect. When a biconvex lens is split into two equal parts by a plane parallel to its principal axis, each part still remains bi-convex, and the focal length will not change. The focal length of each part will remain $10 c m$ , not become $20 c m$ .
Focal length does not depend on the radii of curvature of the two surfaces. This is not correct. The focal length of a lens is directly related to the radii of curvature of its surfaces. The lens formula, which is $1 / f = (n - 1) (1 / R 1 - 1 / R 2),$ clearly shows that the radii of curvature ( $R 1$ and $R 2$ ) affect the focal length $(f)$ of the lens.
Since both the Assertion and the Reason are incorrect, the correct choice is (4).

AIIMS - 2007

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364778 Find the position of final image from first lens. Given, focal length of each lens is $10 c m .$
supporting img

1

40 c m

2

50 c m

3

45 c m

4

55 c m

Explanation:

supporting img
(1) For refraction from 1st lens,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{40} = \frac{3}{40} c m$
$\Rightarrow v = \frac{40}{3} c m$
This image acts as object for 2nd lens.
$∴$ Object distance for 2 nd lens $L_{2}$
$= 30 - \frac{40}{3} = \frac{50}{3} c m$
(2) Now for refraction from $L_{2}$ ,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{10} - \frac{3}{50} = \frac{2}{50}$
$\Rightarrow v = 25 c m$
This image is object for 3rd lens $L_{3}$
$(u = 30 - 25 = 5 c m)$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{5}$
$v = - 10 c m$
(location of final image $= 10 c m$ to
left of $L_{3}$ )
$∴$ Distance from $L_{1}$
$= 30 + (30 - 10) = 50 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364779 The focal length of a convex lens of refractive index 1.5 is $f$ . If it is cut into four equal parts as shown in the figure. The focal length of the one of the parts (Shaded) will be
supporting img

1

f

2

2 f

3

3 f

4

4 f

Explanation:

As $\frac{1}{f^{'}} = (μ - 1) (\frac{1}{R} - \frac{1}{\infty})$
$\frac{1}{f^{'}} = \frac{μ - 1}{R} (1)$
and, $\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R})$
$\frac{1}{f} = \frac{2 (μ - 1)}{R} (2)$
Dividing (2) by (1), we get $f^{'} = 2 f$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364780 A lens is cut from the optical centre along the principal axis and the two parts of the lens are placed sideways as shown in figure. Then the new focal length of the combination:
supporting img

1 Remains the same

2 Becomes double

3 Becomes half

4 Becomes triple

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364781 Assertion :
A bi-convex lens of focal length $10 c m$ is split into two equals parts by a plane parallel to its principal axis. The focal length of the each part will be $20 c m$
Reason :
Focal length does not depend on the radii of curvature of two surfaces.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

A bi-convex lens with a focal length of $10 c m$ is split into two equal parts by a plane parallel to its principal axis, and the focal length of each part will be $20 c m$ . This is incorrect. When a biconvex lens is split into two equal parts by a plane parallel to its principal axis, each part still remains bi-convex, and the focal length will not change. The focal length of each part will remain $10 c m$ , not become $20 c m$ .
Focal length does not depend on the radii of curvature of the two surfaces. This is not correct. The focal length of a lens is directly related to the radii of curvature of its surfaces. The lens formula, which is $1 / f = (n - 1) (1 / R 1 - 1 / R 2),$ clearly shows that the radii of curvature ( $R 1$ and $R 2$ ) affect the focal length $(f)$ of the lens.
Since both the Assertion and the Reason are incorrect, the correct choice is (4).

AIIMS - 2007

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364778 Find the position of final image from first lens. Given, focal length of each lens is $10 c m .$
supporting img

1

40 c m

2

50 c m

3

45 c m

4

55 c m

Explanation:

supporting img
(1) For refraction from 1st lens,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{40} = \frac{3}{40} c m$
$\Rightarrow v = \frac{40}{3} c m$
This image acts as object for 2nd lens.
$∴$ Object distance for 2 nd lens $L_{2}$
$= 30 - \frac{40}{3} = \frac{50}{3} c m$
(2) Now for refraction from $L_{2}$ ,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{10} - \frac{3}{50} = \frac{2}{50}$
$\Rightarrow v = 25 c m$
This image is object for 3rd lens $L_{3}$
$(u = 30 - 25 = 5 c m)$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{5}$
$v = - 10 c m$
(location of final image $= 10 c m$ to
left of $L_{3}$ )
$∴$ Distance from $L_{1}$
$= 30 + (30 - 10) = 50 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364779 The focal length of a convex lens of refractive index 1.5 is $f$ . If it is cut into four equal parts as shown in the figure. The focal length of the one of the parts (Shaded) will be
supporting img

1

f

2

2 f

3

3 f

4

4 f

Explanation:

As $\frac{1}{f^{'}} = (μ - 1) (\frac{1}{R} - \frac{1}{\infty})$
$\frac{1}{f^{'}} = \frac{μ - 1}{R} (1)$
and, $\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R})$
$\frac{1}{f} = \frac{2 (μ - 1)}{R} (2)$
Dividing (2) by (1), we get $f^{'} = 2 f$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364780 A lens is cut from the optical centre along the principal axis and the two parts of the lens are placed sideways as shown in figure. Then the new focal length of the combination:
supporting img

1 Remains the same

2 Becomes double

3 Becomes half

4 Becomes triple

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364781 Assertion :
A bi-convex lens of focal length $10 c m$ is split into two equals parts by a plane parallel to its principal axis. The focal length of the each part will be $20 c m$
Reason :
Focal length does not depend on the radii of curvature of two surfaces.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

A bi-convex lens with a focal length of $10 c m$ is split into two equal parts by a plane parallel to its principal axis, and the focal length of each part will be $20 c m$ . This is incorrect. When a biconvex lens is split into two equal parts by a plane parallel to its principal axis, each part still remains bi-convex, and the focal length will not change. The focal length of each part will remain $10 c m$ , not become $20 c m$ .
Focal length does not depend on the radii of curvature of the two surfaces. This is not correct. The focal length of a lens is directly related to the radii of curvature of its surfaces. The lens formula, which is $1 / f = (n - 1) (1 / R 1 - 1 / R 2),$ clearly shows that the radii of curvature ( $R 1$ and $R 2$ ) affect the focal length $(f)$ of the lens.
Since both the Assertion and the Reason are incorrect, the correct choice is (4).

AIIMS - 2007

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364778 Find the position of final image from first lens. Given, focal length of each lens is $10 c m .$
supporting img

1

40 c m

2

50 c m

3

45 c m

4

55 c m

Explanation:

supporting img
(1) For refraction from 1st lens,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{40} = \frac{3}{40} c m$
$\Rightarrow v = \frac{40}{3} c m$
This image acts as object for 2nd lens.
$∴$ Object distance for 2 nd lens $L_{2}$
$= 30 - \frac{40}{3} = \frac{50}{3} c m$
(2) Now for refraction from $L_{2}$ ,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{10} - \frac{3}{50} = \frac{2}{50}$
$\Rightarrow v = 25 c m$
This image is object for 3rd lens $L_{3}$
$(u = 30 - 25 = 5 c m)$
$\Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{5}$
$v = - 10 c m$
(location of final image $= 10 c m$ to
left of $L_{3}$ )
$∴$ Distance from $L_{1}$
$= 30 + (30 - 10) = 50 c m .$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364779 The focal length of a convex lens of refractive index 1.5 is $f$ . If it is cut into four equal parts as shown in the figure. The focal length of the one of the parts (Shaded) will be
supporting img

1

f

2

2 f

3

3 f

4

4 f

Explanation:

As $\frac{1}{f^{'}} = (μ - 1) (\frac{1}{R} - \frac{1}{\infty})$
$\frac{1}{f^{'}} = \frac{μ - 1}{R} (1)$
and, $\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R})$
$\frac{1}{f} = \frac{2 (μ - 1)}{R} (2)$
Dividing (2) by (1), we get $f^{'} = 2 f$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364780 A lens is cut from the optical centre along the principal axis and the two parts of the lens are placed sideways as shown in figure. Then the new focal length of the combination:
supporting img

1 Remains the same

2 Becomes double

3 Becomes half

4 Becomes triple

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364781 Assertion :
A bi-convex lens of focal length $10 c m$ is split into two equals parts by a plane parallel to its principal axis. The focal length of the each part will be $20 c m$
Reason :
Focal length does not depend on the radii of curvature of two surfaces.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect.

Explanation:

A bi-convex lens with a focal length of $10 c m$ is split into two equal parts by a plane parallel to its principal axis, and the focal length of each part will be $20 c m$ . This is incorrect. When a biconvex lens is split into two equal parts by a plane parallel to its principal axis, each part still remains bi-convex, and the focal length will not change. The focal length of each part will remain $10 c m$ , not become $20 c m$ .
Focal length does not depend on the radii of curvature of the two surfaces. This is not correct. The focal length of a lens is directly related to the radii of curvature of its surfaces. The lens formula, which is $1 / f = (n - 1) (1 / R 1 - 1 / R 2),$ clearly shows that the radii of curvature ( $R 1$ and $R 2$ ) affect the focal length $(f)$ of the lens.
Since both the Assertion and the Reason are incorrect, the correct choice is (4).

AIIMS - 2007