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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364706 An object is placed at $20 c m$ in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?

1

10 c m

2

15 c m

3

7.5 c m

4

6.6 c m

Explanation:

As image is real, magnification
$m = - \frac{v}{u} = - 3 \Rightarrow v = 3 u$
Here, $u = - 20 c m .$
$∴ v = - 60 c m .$
According to mirror formula,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{- 60} + \frac{1}{- 20} = - \frac{4}{60}$
$\Rightarrow f = - 15 c m \Rightarrow | f | = 15 c m$

KCET - 2014

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364707 A square $A B C D$ of side $1 m m$ is kept at distance $15 c m$ infront of the concave mirror as shown in the figure. The focal length of the mirror is $10 c m$ . The length of the perimeter of its image will be (nearly):

1

8 m m

2

2 m m

3

12 m m

4

6 m m

Explanation:

By mirror formula,
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} + \frac{1}{15} = \frac{5}{- 150} = \frac{1}{- 30}$
So, $v = - 30 c m$
Let $A^{'} B^{'}$ and $C^{'} D^{'}$ be image of $A B$ and $C D$ respectively.
$m = - \frac{v}{u} = - 2$
$∴ A^{'} B^{'} = C^{'} D^{'} = 2 \times 1 = 2 m m$
Let $B^{'} C^{'}$ and $A^{'} D^{'}$ be image of $B C$ and $A D$ respectively.
$\frac{B^{'} C^{'}}{B C} = \frac{A^{'} D^{'}}{A D} = \frac{v^{2}}{u^{2}} = 4$
$\Rightarrow B^{'} C^{'} = A^{'} D^{'} = 4 m m$
$∴$ Length $= 2 + 2 + 4 + 4 = 12 m m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364708 A short linear object of length $b$ lies along the axis of a concave mirror. The size of the image is equal to

1

b (\frac{f}{u - f})

2

b {(\frac{u - f}{f})}^{1 / 2}

3

b {(\frac{f}{f - u})}^{2}

4

b (\frac{u - f}{f})

Explanation:

For a spherical mirror
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{- d u}{u^{2}} - \frac{d v}{v^{2}} = 0$
$d v = - {(\frac{v}{u})}^{2} d u = - m^{2} d u$
Where $d u$ and $d v$ are lenghts of small object and its image $d v = - {(\frac{f}{f - u})}^{2} b$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364709 With a concave mirror, an object is placed at a distance $x_{1}$ from the principle focus, on the principle axis. The image is formed at a distance $x_{2}$ from the principle focus. The focal length of the mirror is

1

\sqrt{\frac{x_{1}}{x_{2}}}

2

\sqrt{x_{1} x_{2}}

3

\frac{x_{1} + x_{2}}{2}

4

x_{1} x_{2}

Explanation:

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{f + x_{1}} + \frac{1}{f + x_{2}} = \frac{1}{f}$
$\Rightarrow \frac{f + x_{2} + f + x_{1}}{(f + x_{1}) (f + x_{2})} = \frac{1}{f}$
$f = \sqrt{x_{1} x_{2}}$
This is Newton’s mirror formula.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364706 An object is placed at $20 c m$ in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?

1

10 c m

2

15 c m

3

7.5 c m

4

6.6 c m

Explanation:

As image is real, magnification
$m = - \frac{v}{u} = - 3 \Rightarrow v = 3 u$
Here, $u = - 20 c m .$
$∴ v = - 60 c m .$
According to mirror formula,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{- 60} + \frac{1}{- 20} = - \frac{4}{60}$
$\Rightarrow f = - 15 c m \Rightarrow | f | = 15 c m$

KCET - 2014

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364707 A square $A B C D$ of side $1 m m$ is kept at distance $15 c m$ infront of the concave mirror as shown in the figure. The focal length of the mirror is $10 c m$ . The length of the perimeter of its image will be (nearly):

1

8 m m

2

2 m m

3

12 m m

4

6 m m

Explanation:

By mirror formula,
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} + \frac{1}{15} = \frac{5}{- 150} = \frac{1}{- 30}$
So, $v = - 30 c m$
Let $A^{'} B^{'}$ and $C^{'} D^{'}$ be image of $A B$ and $C D$ respectively.
$m = - \frac{v}{u} = - 2$
$∴ A^{'} B^{'} = C^{'} D^{'} = 2 \times 1 = 2 m m$
Let $B^{'} C^{'}$ and $A^{'} D^{'}$ be image of $B C$ and $A D$ respectively.
$\frac{B^{'} C^{'}}{B C} = \frac{A^{'} D^{'}}{A D} = \frac{v^{2}}{u^{2}} = 4$
$\Rightarrow B^{'} C^{'} = A^{'} D^{'} = 4 m m$
$∴$ Length $= 2 + 2 + 4 + 4 = 12 m m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364708 A short linear object of length $b$ lies along the axis of a concave mirror. The size of the image is equal to

1

b (\frac{f}{u - f})

2

b {(\frac{u - f}{f})}^{1 / 2}

3

b {(\frac{f}{f - u})}^{2}

4

b (\frac{u - f}{f})

Explanation:

For a spherical mirror
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{- d u}{u^{2}} - \frac{d v}{v^{2}} = 0$
$d v = - {(\frac{v}{u})}^{2} d u = - m^{2} d u$
Where $d u$ and $d v$ are lenghts of small object and its image $d v = - {(\frac{f}{f - u})}^{2} b$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364709 With a concave mirror, an object is placed at a distance $x_{1}$ from the principle focus, on the principle axis. The image is formed at a distance $x_{2}$ from the principle focus. The focal length of the mirror is

1

\sqrt{\frac{x_{1}}{x_{2}}}

2

\sqrt{x_{1} x_{2}}

3

\frac{x_{1} + x_{2}}{2}

4

x_{1} x_{2}

Explanation:

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{f + x_{1}} + \frac{1}{f + x_{2}} = \frac{1}{f}$
$\Rightarrow \frac{f + x_{2} + f + x_{1}}{(f + x_{1}) (f + x_{2})} = \frac{1}{f}$
$f = \sqrt{x_{1} x_{2}}$
This is Newton’s mirror formula.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364706 An object is placed at $20 c m$ in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?

1

10 c m

2

15 c m

3

7.5 c m

4

6.6 c m

Explanation:

As image is real, magnification
$m = - \frac{v}{u} = - 3 \Rightarrow v = 3 u$
Here, $u = - 20 c m .$
$∴ v = - 60 c m .$
According to mirror formula,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{- 60} + \frac{1}{- 20} = - \frac{4}{60}$
$\Rightarrow f = - 15 c m \Rightarrow | f | = 15 c m$

KCET - 2014

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364707 A square $A B C D$ of side $1 m m$ is kept at distance $15 c m$ infront of the concave mirror as shown in the figure. The focal length of the mirror is $10 c m$ . The length of the perimeter of its image will be (nearly):

1

8 m m

2

2 m m

3

12 m m

4

6 m m

Explanation:

By mirror formula,
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} + \frac{1}{15} = \frac{5}{- 150} = \frac{1}{- 30}$
So, $v = - 30 c m$
Let $A^{'} B^{'}$ and $C^{'} D^{'}$ be image of $A B$ and $C D$ respectively.
$m = - \frac{v}{u} = - 2$
$∴ A^{'} B^{'} = C^{'} D^{'} = 2 \times 1 = 2 m m$
Let $B^{'} C^{'}$ and $A^{'} D^{'}$ be image of $B C$ and $A D$ respectively.
$\frac{B^{'} C^{'}}{B C} = \frac{A^{'} D^{'}}{A D} = \frac{v^{2}}{u^{2}} = 4$
$\Rightarrow B^{'} C^{'} = A^{'} D^{'} = 4 m m$
$∴$ Length $= 2 + 2 + 4 + 4 = 12 m m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364708 A short linear object of length $b$ lies along the axis of a concave mirror. The size of the image is equal to

1

b (\frac{f}{u - f})

2

b {(\frac{u - f}{f})}^{1 / 2}

3

b {(\frac{f}{f - u})}^{2}

4

b (\frac{u - f}{f})

Explanation:

For a spherical mirror
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{- d u}{u^{2}} - \frac{d v}{v^{2}} = 0$
$d v = - {(\frac{v}{u})}^{2} d u = - m^{2} d u$
Where $d u$ and $d v$ are lenghts of small object and its image $d v = - {(\frac{f}{f - u})}^{2} b$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364709 With a concave mirror, an object is placed at a distance $x_{1}$ from the principle focus, on the principle axis. The image is formed at a distance $x_{2}$ from the principle focus. The focal length of the mirror is

1

\sqrt{\frac{x_{1}}{x_{2}}}

2

\sqrt{x_{1} x_{2}}

3

\frac{x_{1} + x_{2}}{2}

4

x_{1} x_{2}

Explanation:

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{f + x_{1}} + \frac{1}{f + x_{2}} = \frac{1}{f}$
$\Rightarrow \frac{f + x_{2} + f + x_{1}}{(f + x_{1}) (f + x_{2})} = \frac{1}{f}$
$f = \sqrt{x_{1} x_{2}}$
This is Newton’s mirror formula.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364706 An object is placed at $20 c m$ in front of a concave mirror produces three times magnified real image. What its focal length of the concave mirror?

1

10 c m

2

15 c m

3

7.5 c m

4

6.6 c m

Explanation:

As image is real, magnification
$m = - \frac{v}{u} = - 3 \Rightarrow v = 3 u$
Here, $u = - 20 c m .$
$∴ v = - 60 c m .$
According to mirror formula,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{- 60} + \frac{1}{- 20} = - \frac{4}{60}$
$\Rightarrow f = - 15 c m \Rightarrow | f | = 15 c m$

KCET - 2014

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364707 A square $A B C D$ of side $1 m m$ is kept at distance $15 c m$ infront of the concave mirror as shown in the figure. The focal length of the mirror is $10 c m$ . The length of the perimeter of its image will be (nearly):

1

8 m m

2

2 m m

3

12 m m

4

6 m m

Explanation:

By mirror formula,
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} + \frac{1}{15} = \frac{5}{- 150} = \frac{1}{- 30}$
So, $v = - 30 c m$
Let $A^{'} B^{'}$ and $C^{'} D^{'}$ be image of $A B$ and $C D$ respectively.
$m = - \frac{v}{u} = - 2$
$∴ A^{'} B^{'} = C^{'} D^{'} = 2 \times 1 = 2 m m$
Let $B^{'} C^{'}$ and $A^{'} D^{'}$ be image of $B C$ and $A D$ respectively.
$\frac{B^{'} C^{'}}{B C} = \frac{A^{'} D^{'}}{A D} = \frac{v^{2}}{u^{2}} = 4$
$\Rightarrow B^{'} C^{'} = A^{'} D^{'} = 4 m m$
$∴$ Length $= 2 + 2 + 4 + 4 = 12 m m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364708 A short linear object of length $b$ lies along the axis of a concave mirror. The size of the image is equal to

1

b (\frac{f}{u - f})

2

b {(\frac{u - f}{f})}^{1 / 2}

3

b {(\frac{f}{f - u})}^{2}

4

b (\frac{u - f}{f})

Explanation:

For a spherical mirror
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{- d u}{u^{2}} - \frac{d v}{v^{2}} = 0$
$d v = - {(\frac{v}{u})}^{2} d u = - m^{2} d u$
Where $d u$ and $d v$ are lenghts of small object and its image $d v = - {(\frac{f}{f - u})}^{2} b$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364709 With a concave mirror, an object is placed at a distance $x_{1}$ from the principle focus, on the principle axis. The image is formed at a distance $x_{2}$ from the principle focus. The focal length of the mirror is

1

\sqrt{\frac{x_{1}}{x_{2}}}

2

\sqrt{x_{1} x_{2}}

3

\frac{x_{1} + x_{2}}{2}

4

x_{1} x_{2}

Explanation:

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{f + x_{1}} + \frac{1}{f + x_{2}} = \frac{1}{f}$
$\Rightarrow \frac{f + x_{2} + f + x_{1}}{(f + x_{1}) (f + x_{2})} = \frac{1}{f}$
$f = \sqrt{x_{1} x_{2}}$
This is Newton’s mirror formula.

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