QBManager

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364635 Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint $M$ of mirror $A D$ and after two reflections reaches corner $D$ . Then, $\cot θ =$
supporting img

1 0.45

2 0.75

3 0.36

4 0.96

Explanation:

The ray starting from point $M$ at an angle $θ$ reaches the corner $D$ at the right along a parallel path. Let a be the length of the side and let light ray strikes edge $A B$ at distance $x$ .
supporting img
From figure,
$\tan θ = \frac{x}{(a / 2)}$ $(1)$
Also, $\tan θ = \frac{a - x}{y}$ $(2)$
and $\tan θ = \frac{a}{a - y}$ $(3)$
From equations (1) and (2), we get
$\frac{2 x}{a} = \frac{a - x}{y}$
$∴ 2 x y = a^{2} - x a (4)$
From equations (2) and (3), we get
$\frac{a - x}{y} = \frac{a}{a - y}$
$∴ a^{2} - x a + x y = 2 a y$
$∴ 3 x y = 2 a y [F r o m (4)]$
$∴ x = \frac{2 a}{3}$
Substituting this value of $x$ in equation (1), we get,
$\tan θ = \frac{(2 a / 3)}{(a / 2)} = \frac{4}{3}$
$∴ \cot θ = \frac{1}{\tan θ} = \frac{3}{4} = 0.75$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364636 A plane mirror is approaching you at a speed of $10 c m / \sec$ . You can see your image in it. At what speed will your image approach you?

1

15 c m / \sec

2

5 c m / \sec

3

10 c m / \sec

4

20 c m / \sec

Explanation:

$V_{I m} = - V_{O m}$
Given that ${\vec{V}}_{O} = 0, {\vec{V}}_{m} = - 10 c m / s \hat{i}$
${\vec{V}}_{I} - {\vec{V}}_{m} = - ({\vec{V}}_{0} - {\vec{V}}_{m}) = {\vec{V}}_{m}$
${\vec{V}}_{I} = 2 {\vec{V}}_{m} = - 20 c m / s \hat{i}$
${\vec{V}}_{I O} = {\vec{V}}_{I} - {\vec{V}}_{0} = - 20 c m / s \hat{i} - 0 = - 20 c m / s \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364637 A plane mirror is placed in $y - z$ plane facing towards negative $x$ -axis. The mirror is moving parallel to $y$ -axis with a speed of $2 c m / s$ . A point object $P$ is moving infront of the mirror with a velocity $(1 c m / s)$ $\hat{i} + (1 c m / s) \hat{j} .$ Then which of the following statements is incorrect?

1 The velocity of image with respect to mirror is

- (\hat{i} + \hat{j}) c m / s

2 The velocity of image is

- \hat{i} + \hat{j} c m / s

3 The unit vector parallel to the plane of the mirror is

\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}

4 The velocity of image with respect to object is

(- 2 \hat{i}) c m / s

Explanation:

Given that
${\vec{V}}_{m} = 2 \hat{j}, {\vec{V}}_{O} = \hat{i} + \hat{j}$
${\vec{V}}_{O m} = {\vec{V}}_{O} - {\vec{V}}_{m} = \hat{i} - \hat{j}$
${\vec{V}}_{I m} = {\vec{V}}_{O m}$ (component parallel to the plane of the mirror)
$V_{Im} = - V_{O m}$ (component perpendicular to the plane of the mirror) $\Rightarrow {\vec{V}}_{I m} = - \hat{i} - \hat{j} \Rightarrow {\vec{V}}_{I} = - \hat{i} + \hat{j}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364638 A person runs with a speed $u$ towards a bicycle moving away from him with speed $v$ . The person approaches his image in the plane of the mirror fixed at the rear of bicycle with a speed of

1

u - 2 v

2

u - v

3

2 (u - v)

4

2 u - v

Explanation:

Assume that both object and mirror are moving towards (+) ve $x$ -axis.
${\vec{v}}_{o} = u \hat{i} {\vec{v}}_{m} = v \hat{i}$
${\vec{v}}_{o m} = {\vec{v}}_{o} - {\vec{v}}_{m} = (u - v) \hat{i}$
$\Rightarrow {\vec{v}}_{I m} = - (u - v) \hat{i}$
${\vec{v}}_{I m} = {\vec{v}}_{I} - {\vec{v}}_{m}$
$\Rightarrow {\vec{v}}_{I} = (- u + 2 v) \hat{i}$
${\vec{v}}_{o I} = {\vec{v}}_{o} - {\vec{v}}_{I} = 2 u \hat{i} - 2 v \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364639 An object is approaching a plane mirror at $10 c m / s .$ A stationary observer sees the image. At what speed will the image approach the stationary observer?

1

15 c m s^{- 1}

2

10 c m s^{- 1}

3

5 c m s^{- 1}

4

20 c m s^{- 1}

Explanation:

$V_{Im} = - V_{O m}$
Given that $V_{m} = 0, V_{O} = 10 c m / s$
$\Rightarrow V_{I} = - V_{O} = - 10 c m / s$
Here $V_{O}$ - velocity of object. Now the second observer who is at rest sees the image moving with velocity $10 c m / s$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364635 Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint $M$ of mirror $A D$ and after two reflections reaches corner $D$ . Then, $\cot θ =$
supporting img

1 0.45

2 0.75

3 0.36

4 0.96

Explanation:

The ray starting from point $M$ at an angle $θ$ reaches the corner $D$ at the right along a parallel path. Let a be the length of the side and let light ray strikes edge $A B$ at distance $x$ .
supporting img
From figure,
$\tan θ = \frac{x}{(a / 2)}$ $(1)$
Also, $\tan θ = \frac{a - x}{y}$ $(2)$
and $\tan θ = \frac{a}{a - y}$ $(3)$
From equations (1) and (2), we get
$\frac{2 x}{a} = \frac{a - x}{y}$
$∴ 2 x y = a^{2} - x a (4)$
From equations (2) and (3), we get
$\frac{a - x}{y} = \frac{a}{a - y}$
$∴ a^{2} - x a + x y = 2 a y$
$∴ 3 x y = 2 a y [F r o m (4)]$
$∴ x = \frac{2 a}{3}$
Substituting this value of $x$ in equation (1), we get,
$\tan θ = \frac{(2 a / 3)}{(a / 2)} = \frac{4}{3}$
$∴ \cot θ = \frac{1}{\tan θ} = \frac{3}{4} = 0.75$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364636 A plane mirror is approaching you at a speed of $10 c m / \sec$ . You can see your image in it. At what speed will your image approach you?

1

15 c m / \sec

2

5 c m / \sec

3

10 c m / \sec

4

20 c m / \sec

Explanation:

$V_{I m} = - V_{O m}$
Given that ${\vec{V}}_{O} = 0, {\vec{V}}_{m} = - 10 c m / s \hat{i}$
${\vec{V}}_{I} - {\vec{V}}_{m} = - ({\vec{V}}_{0} - {\vec{V}}_{m}) = {\vec{V}}_{m}$
${\vec{V}}_{I} = 2 {\vec{V}}_{m} = - 20 c m / s \hat{i}$
${\vec{V}}_{I O} = {\vec{V}}_{I} - {\vec{V}}_{0} = - 20 c m / s \hat{i} - 0 = - 20 c m / s \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364637 A plane mirror is placed in $y - z$ plane facing towards negative $x$ -axis. The mirror is moving parallel to $y$ -axis with a speed of $2 c m / s$ . A point object $P$ is moving infront of the mirror with a velocity $(1 c m / s)$ $\hat{i} + (1 c m / s) \hat{j} .$ Then which of the following statements is incorrect?

1 The velocity of image with respect to mirror is

- (\hat{i} + \hat{j}) c m / s

2 The velocity of image is

- \hat{i} + \hat{j} c m / s

3 The unit vector parallel to the plane of the mirror is

\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}

4 The velocity of image with respect to object is

(- 2 \hat{i}) c m / s

Explanation:

Given that
${\vec{V}}_{m} = 2 \hat{j}, {\vec{V}}_{O} = \hat{i} + \hat{j}$
${\vec{V}}_{O m} = {\vec{V}}_{O} - {\vec{V}}_{m} = \hat{i} - \hat{j}$
${\vec{V}}_{I m} = {\vec{V}}_{O m}$ (component parallel to the plane of the mirror)
$V_{Im} = - V_{O m}$ (component perpendicular to the plane of the mirror) $\Rightarrow {\vec{V}}_{I m} = - \hat{i} - \hat{j} \Rightarrow {\vec{V}}_{I} = - \hat{i} + \hat{j}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364638 A person runs with a speed $u$ towards a bicycle moving away from him with speed $v$ . The person approaches his image in the plane of the mirror fixed at the rear of bicycle with a speed of

1

u - 2 v

2

u - v

3

2 (u - v)

4

2 u - v

Explanation:

Assume that both object and mirror are moving towards (+) ve $x$ -axis.
${\vec{v}}_{o} = u \hat{i} {\vec{v}}_{m} = v \hat{i}$
${\vec{v}}_{o m} = {\vec{v}}_{o} - {\vec{v}}_{m} = (u - v) \hat{i}$
$\Rightarrow {\vec{v}}_{I m} = - (u - v) \hat{i}$
${\vec{v}}_{I m} = {\vec{v}}_{I} - {\vec{v}}_{m}$
$\Rightarrow {\vec{v}}_{I} = (- u + 2 v) \hat{i}$
${\vec{v}}_{o I} = {\vec{v}}_{o} - {\vec{v}}_{I} = 2 u \hat{i} - 2 v \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364639 An object is approaching a plane mirror at $10 c m / s .$ A stationary observer sees the image. At what speed will the image approach the stationary observer?

1

15 c m s^{- 1}

2

10 c m s^{- 1}

3

5 c m s^{- 1}

4

20 c m s^{- 1}

Explanation:

$V_{Im} = - V_{O m}$
Given that $V_{m} = 0, V_{O} = 10 c m / s$
$\Rightarrow V_{I} = - V_{O} = - 10 c m / s$
Here $V_{O}$ - velocity of object. Now the second observer who is at rest sees the image moving with velocity $10 c m / s$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364635 Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint $M$ of mirror $A D$ and after two reflections reaches corner $D$ . Then, $\cot θ =$
supporting img

1 0.45

2 0.75

3 0.36

4 0.96

Explanation:

The ray starting from point $M$ at an angle $θ$ reaches the corner $D$ at the right along a parallel path. Let a be the length of the side and let light ray strikes edge $A B$ at distance $x$ .
supporting img
From figure,
$\tan θ = \frac{x}{(a / 2)}$ $(1)$
Also, $\tan θ = \frac{a - x}{y}$ $(2)$
and $\tan θ = \frac{a}{a - y}$ $(3)$
From equations (1) and (2), we get
$\frac{2 x}{a} = \frac{a - x}{y}$
$∴ 2 x y = a^{2} - x a (4)$
From equations (2) and (3), we get
$\frac{a - x}{y} = \frac{a}{a - y}$
$∴ a^{2} - x a + x y = 2 a y$
$∴ 3 x y = 2 a y [F r o m (4)]$
$∴ x = \frac{2 a}{3}$
Substituting this value of $x$ in equation (1), we get,
$\tan θ = \frac{(2 a / 3)}{(a / 2)} = \frac{4}{3}$
$∴ \cot θ = \frac{1}{\tan θ} = \frac{3}{4} = 0.75$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364636 A plane mirror is approaching you at a speed of $10 c m / \sec$ . You can see your image in it. At what speed will your image approach you?

1

15 c m / \sec

2

5 c m / \sec

3

10 c m / \sec

4

20 c m / \sec

Explanation:

$V_{I m} = - V_{O m}$
Given that ${\vec{V}}_{O} = 0, {\vec{V}}_{m} = - 10 c m / s \hat{i}$
${\vec{V}}_{I} - {\vec{V}}_{m} = - ({\vec{V}}_{0} - {\vec{V}}_{m}) = {\vec{V}}_{m}$
${\vec{V}}_{I} = 2 {\vec{V}}_{m} = - 20 c m / s \hat{i}$
${\vec{V}}_{I O} = {\vec{V}}_{I} - {\vec{V}}_{0} = - 20 c m / s \hat{i} - 0 = - 20 c m / s \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364637 A plane mirror is placed in $y - z$ plane facing towards negative $x$ -axis. The mirror is moving parallel to $y$ -axis with a speed of $2 c m / s$ . A point object $P$ is moving infront of the mirror with a velocity $(1 c m / s)$ $\hat{i} + (1 c m / s) \hat{j} .$ Then which of the following statements is incorrect?

1 The velocity of image with respect to mirror is

- (\hat{i} + \hat{j}) c m / s

2 The velocity of image is

- \hat{i} + \hat{j} c m / s

3 The unit vector parallel to the plane of the mirror is

\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}

4 The velocity of image with respect to object is

(- 2 \hat{i}) c m / s

Explanation:

Given that
${\vec{V}}_{m} = 2 \hat{j}, {\vec{V}}_{O} = \hat{i} + \hat{j}$
${\vec{V}}_{O m} = {\vec{V}}_{O} - {\vec{V}}_{m} = \hat{i} - \hat{j}$
${\vec{V}}_{I m} = {\vec{V}}_{O m}$ (component parallel to the plane of the mirror)
$V_{Im} = - V_{O m}$ (component perpendicular to the plane of the mirror) $\Rightarrow {\vec{V}}_{I m} = - \hat{i} - \hat{j} \Rightarrow {\vec{V}}_{I} = - \hat{i} + \hat{j}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364638 A person runs with a speed $u$ towards a bicycle moving away from him with speed $v$ . The person approaches his image in the plane of the mirror fixed at the rear of bicycle with a speed of

1

u - 2 v

2

u - v

3

2 (u - v)

4

2 u - v

Explanation:

Assume that both object and mirror are moving towards (+) ve $x$ -axis.
${\vec{v}}_{o} = u \hat{i} {\vec{v}}_{m} = v \hat{i}$
${\vec{v}}_{o m} = {\vec{v}}_{o} - {\vec{v}}_{m} = (u - v) \hat{i}$
$\Rightarrow {\vec{v}}_{I m} = - (u - v) \hat{i}$
${\vec{v}}_{I m} = {\vec{v}}_{I} - {\vec{v}}_{m}$
$\Rightarrow {\vec{v}}_{I} = (- u + 2 v) \hat{i}$
${\vec{v}}_{o I} = {\vec{v}}_{o} - {\vec{v}}_{I} = 2 u \hat{i} - 2 v \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364639 An object is approaching a plane mirror at $10 c m / s .$ A stationary observer sees the image. At what speed will the image approach the stationary observer?

1

15 c m s^{- 1}

2

10 c m s^{- 1}

3

5 c m s^{- 1}

4

20 c m s^{- 1}

Explanation:

$V_{Im} = - V_{O m}$
Given that $V_{m} = 0, V_{O} = 10 c m / s$
$\Rightarrow V_{I} = - V_{O} = - 10 c m / s$
Here $V_{O}$ - velocity of object. Now the second observer who is at rest sees the image moving with velocity $10 c m / s$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364635 Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint $M$ of mirror $A D$ and after two reflections reaches corner $D$ . Then, $\cot θ =$
supporting img

1 0.45

2 0.75

3 0.36

4 0.96

Explanation:

The ray starting from point $M$ at an angle $θ$ reaches the corner $D$ at the right along a parallel path. Let a be the length of the side and let light ray strikes edge $A B$ at distance $x$ .
supporting img
From figure,
$\tan θ = \frac{x}{(a / 2)}$ $(1)$
Also, $\tan θ = \frac{a - x}{y}$ $(2)$
and $\tan θ = \frac{a}{a - y}$ $(3)$
From equations (1) and (2), we get
$\frac{2 x}{a} = \frac{a - x}{y}$
$∴ 2 x y = a^{2} - x a (4)$
From equations (2) and (3), we get
$\frac{a - x}{y} = \frac{a}{a - y}$
$∴ a^{2} - x a + x y = 2 a y$
$∴ 3 x y = 2 a y [F r o m (4)]$
$∴ x = \frac{2 a}{3}$
Substituting this value of $x$ in equation (1), we get,
$\tan θ = \frac{(2 a / 3)}{(a / 2)} = \frac{4}{3}$
$∴ \cot θ = \frac{1}{\tan θ} = \frac{3}{4} = 0.75$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364636 A plane mirror is approaching you at a speed of $10 c m / \sec$ . You can see your image in it. At what speed will your image approach you?

1

15 c m / \sec

2

5 c m / \sec

3

10 c m / \sec

4

20 c m / \sec

Explanation:

$V_{I m} = - V_{O m}$
Given that ${\vec{V}}_{O} = 0, {\vec{V}}_{m} = - 10 c m / s \hat{i}$
${\vec{V}}_{I} - {\vec{V}}_{m} = - ({\vec{V}}_{0} - {\vec{V}}_{m}) = {\vec{V}}_{m}$
${\vec{V}}_{I} = 2 {\vec{V}}_{m} = - 20 c m / s \hat{i}$
${\vec{V}}_{I O} = {\vec{V}}_{I} - {\vec{V}}_{0} = - 20 c m / s \hat{i} - 0 = - 20 c m / s \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364637 A plane mirror is placed in $y - z$ plane facing towards negative $x$ -axis. The mirror is moving parallel to $y$ -axis with a speed of $2 c m / s$ . A point object $P$ is moving infront of the mirror with a velocity $(1 c m / s)$ $\hat{i} + (1 c m / s) \hat{j} .$ Then which of the following statements is incorrect?

1 The velocity of image with respect to mirror is

- (\hat{i} + \hat{j}) c m / s

2 The velocity of image is

- \hat{i} + \hat{j} c m / s

3 The unit vector parallel to the plane of the mirror is

\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}

4 The velocity of image with respect to object is

(- 2 \hat{i}) c m / s

Explanation:

Given that
${\vec{V}}_{m} = 2 \hat{j}, {\vec{V}}_{O} = \hat{i} + \hat{j}$
${\vec{V}}_{O m} = {\vec{V}}_{O} - {\vec{V}}_{m} = \hat{i} - \hat{j}$
${\vec{V}}_{I m} = {\vec{V}}_{O m}$ (component parallel to the plane of the mirror)
$V_{Im} = - V_{O m}$ (component perpendicular to the plane of the mirror) $\Rightarrow {\vec{V}}_{I m} = - \hat{i} - \hat{j} \Rightarrow {\vec{V}}_{I} = - \hat{i} + \hat{j}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364638 A person runs with a speed $u$ towards a bicycle moving away from him with speed $v$ . The person approaches his image in the plane of the mirror fixed at the rear of bicycle with a speed of

1

u - 2 v

2

u - v

3

2 (u - v)

4

2 u - v

Explanation:

Assume that both object and mirror are moving towards (+) ve $x$ -axis.
${\vec{v}}_{o} = u \hat{i} {\vec{v}}_{m} = v \hat{i}$
${\vec{v}}_{o m} = {\vec{v}}_{o} - {\vec{v}}_{m} = (u - v) \hat{i}$
$\Rightarrow {\vec{v}}_{I m} = - (u - v) \hat{i}$
${\vec{v}}_{I m} = {\vec{v}}_{I} - {\vec{v}}_{m}$
$\Rightarrow {\vec{v}}_{I} = (- u + 2 v) \hat{i}$
${\vec{v}}_{o I} = {\vec{v}}_{o} - {\vec{v}}_{I} = 2 u \hat{i} - 2 v \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364639 An object is approaching a plane mirror at $10 c m / s .$ A stationary observer sees the image. At what speed will the image approach the stationary observer?

1

15 c m s^{- 1}

2

10 c m s^{- 1}

3

5 c m s^{- 1}

4

20 c m s^{- 1}

Explanation:

$V_{Im} = - V_{O m}$
Given that $V_{m} = 0, V_{O} = 10 c m / s$
$\Rightarrow V_{I} = - V_{O} = - 10 c m / s$
Here $V_{O}$ - velocity of object. Now the second observer who is at rest sees the image moving with velocity $10 c m / s$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364635 Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint $M$ of mirror $A D$ and after two reflections reaches corner $D$ . Then, $\cot θ =$
supporting img

1 0.45

2 0.75

3 0.36

4 0.96

Explanation:

The ray starting from point $M$ at an angle $θ$ reaches the corner $D$ at the right along a parallel path. Let a be the length of the side and let light ray strikes edge $A B$ at distance $x$ .
supporting img
From figure,
$\tan θ = \frac{x}{(a / 2)}$ $(1)$
Also, $\tan θ = \frac{a - x}{y}$ $(2)$
and $\tan θ = \frac{a}{a - y}$ $(3)$
From equations (1) and (2), we get
$\frac{2 x}{a} = \frac{a - x}{y}$
$∴ 2 x y = a^{2} - x a (4)$
From equations (2) and (3), we get
$\frac{a - x}{y} = \frac{a}{a - y}$
$∴ a^{2} - x a + x y = 2 a y$
$∴ 3 x y = 2 a y [F r o m (4)]$
$∴ x = \frac{2 a}{3}$
Substituting this value of $x$ in equation (1), we get,
$\tan θ = \frac{(2 a / 3)}{(a / 2)} = \frac{4}{3}$
$∴ \cot θ = \frac{1}{\tan θ} = \frac{3}{4} = 0.75$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364636 A plane mirror is approaching you at a speed of $10 c m / \sec$ . You can see your image in it. At what speed will your image approach you?

1

15 c m / \sec

2

5 c m / \sec

3

10 c m / \sec

4

20 c m / \sec

Explanation:

$V_{I m} = - V_{O m}$
Given that ${\vec{V}}_{O} = 0, {\vec{V}}_{m} = - 10 c m / s \hat{i}$
${\vec{V}}_{I} - {\vec{V}}_{m} = - ({\vec{V}}_{0} - {\vec{V}}_{m}) = {\vec{V}}_{m}$
${\vec{V}}_{I} = 2 {\vec{V}}_{m} = - 20 c m / s \hat{i}$
${\vec{V}}_{I O} = {\vec{V}}_{I} - {\vec{V}}_{0} = - 20 c m / s \hat{i} - 0 = - 20 c m / s \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364637 A plane mirror is placed in $y - z$ plane facing towards negative $x$ -axis. The mirror is moving parallel to $y$ -axis with a speed of $2 c m / s$ . A point object $P$ is moving infront of the mirror with a velocity $(1 c m / s)$ $\hat{i} + (1 c m / s) \hat{j} .$ Then which of the following statements is incorrect?

1 The velocity of image with respect to mirror is

- (\hat{i} + \hat{j}) c m / s

2 The velocity of image is

- \hat{i} + \hat{j} c m / s

3 The unit vector parallel to the plane of the mirror is

\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}

4 The velocity of image with respect to object is

(- 2 \hat{i}) c m / s

Explanation:

Given that
${\vec{V}}_{m} = 2 \hat{j}, {\vec{V}}_{O} = \hat{i} + \hat{j}$
${\vec{V}}_{O m} = {\vec{V}}_{O} - {\vec{V}}_{m} = \hat{i} - \hat{j}$
${\vec{V}}_{I m} = {\vec{V}}_{O m}$ (component parallel to the plane of the mirror)
$V_{Im} = - V_{O m}$ (component perpendicular to the plane of the mirror) $\Rightarrow {\vec{V}}_{I m} = - \hat{i} - \hat{j} \Rightarrow {\vec{V}}_{I} = - \hat{i} + \hat{j}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364638 A person runs with a speed $u$ towards a bicycle moving away from him with speed $v$ . The person approaches his image in the plane of the mirror fixed at the rear of bicycle with a speed of

1

u - 2 v

2

u - v

3

2 (u - v)

4

2 u - v

Explanation:

Assume that both object and mirror are moving towards (+) ve $x$ -axis.
${\vec{v}}_{o} = u \hat{i} {\vec{v}}_{m} = v \hat{i}$
${\vec{v}}_{o m} = {\vec{v}}_{o} - {\vec{v}}_{m} = (u - v) \hat{i}$
$\Rightarrow {\vec{v}}_{I m} = - (u - v) \hat{i}$
${\vec{v}}_{I m} = {\vec{v}}_{I} - {\vec{v}}_{m}$
$\Rightarrow {\vec{v}}_{I} = (- u + 2 v) \hat{i}$
${\vec{v}}_{o I} = {\vec{v}}_{o} - {\vec{v}}_{I} = 2 u \hat{i} - 2 v \hat{i}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364639 An object is approaching a plane mirror at $10 c m / s .$ A stationary observer sees the image. At what speed will the image approach the stationary observer?

1

15 c m s^{- 1}

2

10 c m s^{- 1}

3

5 c m s^{- 1}

4

20 c m s^{- 1}

Explanation:

$V_{Im} = - V_{O m}$
Given that $V_{m} = 0, V_{O} = 10 c m / s$
$\Rightarrow V_{I} = - V_{O} = - 10 c m / s$
Here $V_{O}$ - velocity of object. Now the second observer who is at rest sees the image moving with velocity $10 c m / s$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: