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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364898 An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $12 c m$ . When viewed from the opposite side, the apparent distance of the bubble is observed as $4 c m$ . If the side of the ice cube is $24 c m$ , the refractive index of the ice cube is

1

\frac{4}{3}

2

\frac{6}{5}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

supporting img
Apparent depth $= ($ real depth $) / μ$ .
From one side
$x / μ = 12 c m (1)$
From opposite side
$(24 - x) / μ = 4 c m (2)$
Divide (1) by (2), we get
$\frac{x / μ}{(24 - x) / μ} = \frac{12}{4} \Rightarrow \frac{x}{24 - x} = 3 \Rightarrow x = 18$
$μ = \frac{18}{12} = \frac{3}{2}$

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364899 A coin is present on the bottom of a water tank of depth $30 c m .$ If the refractive index of the water is 1.5 then shift in the position of the coin as seen by an observer standing in air is

1

10 c m

2

20 c m

3

5 c m

4

2 c m

Explanation:

Shift $= d (1 - \frac{1}{μ}) = 30 (1 - \frac{1}{1.5}) = 10 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364900 When the rectangular metal tank is filled up to the top with an unknown liquid, an observer with eyes level with the top of the tank can just see the corner $E$ . Find the index of refraction of the liquid
supporting img

1 1.5

2 1.25

3 1.67

4 2.4

Explanation:

In $△ A B C$
$Sin θ_{c} = \frac{60}{\sqrt{60^{2} + 80^{2}}} = \frac{60}{100}$

supporting img

Also, $\sin θ_{C} = \frac{1}{μ}$
$\Rightarrow \frac{60}{100} = \frac{1}{μ} [∵ Sin θ c = \frac{100}{60}] \Rightarrow μ = 1.67$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364901 A fish at a depth of $12 c m$ in water $(μ = \frac{3}{4})$ is viewed by an observer on the bank of a lake. Its apparent rise as observed by the observer

1

3 c m

2

9 c m

3

8 c m

4

4 c m

Explanation:

Apparent rise
$= d (1 - \frac{1}{μ}) = 12 \times (1 - \frac{3}{4}) = 3 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364898 An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $12 c m$ . When viewed from the opposite side, the apparent distance of the bubble is observed as $4 c m$ . If the side of the ice cube is $24 c m$ , the refractive index of the ice cube is

1

\frac{4}{3}

2

\frac{6}{5}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

supporting img
Apparent depth $= ($ real depth $) / μ$ .
From one side
$x / μ = 12 c m (1)$
From opposite side
$(24 - x) / μ = 4 c m (2)$
Divide (1) by (2), we get
$\frac{x / μ}{(24 - x) / μ} = \frac{12}{4} \Rightarrow \frac{x}{24 - x} = 3 \Rightarrow x = 18$
$μ = \frac{18}{12} = \frac{3}{2}$

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364899 A coin is present on the bottom of a water tank of depth $30 c m .$ If the refractive index of the water is 1.5 then shift in the position of the coin as seen by an observer standing in air is

1

10 c m

2

20 c m

3

5 c m

4

2 c m

Explanation:

Shift $= d (1 - \frac{1}{μ}) = 30 (1 - \frac{1}{1.5}) = 10 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364900 When the rectangular metal tank is filled up to the top with an unknown liquid, an observer with eyes level with the top of the tank can just see the corner $E$ . Find the index of refraction of the liquid
supporting img

1 1.5

2 1.25

3 1.67

4 2.4

Explanation:

In $△ A B C$
$Sin θ_{c} = \frac{60}{\sqrt{60^{2} + 80^{2}}} = \frac{60}{100}$

supporting img

Also, $\sin θ_{C} = \frac{1}{μ}$
$\Rightarrow \frac{60}{100} = \frac{1}{μ} [∵ Sin θ c = \frac{100}{60}] \Rightarrow μ = 1.67$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364901 A fish at a depth of $12 c m$ in water $(μ = \frac{3}{4})$ is viewed by an observer on the bank of a lake. Its apparent rise as observed by the observer

1

3 c m

2

9 c m

3

8 c m

4

4 c m

Explanation:

Apparent rise
$= d (1 - \frac{1}{μ}) = 12 \times (1 - \frac{3}{4}) = 3 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364898 An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $12 c m$ . When viewed from the opposite side, the apparent distance of the bubble is observed as $4 c m$ . If the side of the ice cube is $24 c m$ , the refractive index of the ice cube is

1

\frac{4}{3}

2

\frac{6}{5}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

supporting img
Apparent depth $= ($ real depth $) / μ$ .
From one side
$x / μ = 12 c m (1)$
From opposite side
$(24 - x) / μ = 4 c m (2)$
Divide (1) by (2), we get
$\frac{x / μ}{(24 - x) / μ} = \frac{12}{4} \Rightarrow \frac{x}{24 - x} = 3 \Rightarrow x = 18$
$μ = \frac{18}{12} = \frac{3}{2}$

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364899 A coin is present on the bottom of a water tank of depth $30 c m .$ If the refractive index of the water is 1.5 then shift in the position of the coin as seen by an observer standing in air is

1

10 c m

2

20 c m

3

5 c m

4

2 c m

Explanation:

Shift $= d (1 - \frac{1}{μ}) = 30 (1 - \frac{1}{1.5}) = 10 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364900 When the rectangular metal tank is filled up to the top with an unknown liquid, an observer with eyes level with the top of the tank can just see the corner $E$ . Find the index of refraction of the liquid
supporting img

1 1.5

2 1.25

3 1.67

4 2.4

Explanation:

In $△ A B C$
$Sin θ_{c} = \frac{60}{\sqrt{60^{2} + 80^{2}}} = \frac{60}{100}$

supporting img

Also, $\sin θ_{C} = \frac{1}{μ}$
$\Rightarrow \frac{60}{100} = \frac{1}{μ} [∵ Sin θ c = \frac{100}{60}] \Rightarrow μ = 1.67$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364901 A fish at a depth of $12 c m$ in water $(μ = \frac{3}{4})$ is viewed by an observer on the bank of a lake. Its apparent rise as observed by the observer

1

3 c m

2

9 c m

3

8 c m

4

4 c m

Explanation:

Apparent rise
$= d (1 - \frac{1}{μ}) = 12 \times (1 - \frac{3}{4}) = 3 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364898 An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $12 c m$ . When viewed from the opposite side, the apparent distance of the bubble is observed as $4 c m$ . If the side of the ice cube is $24 c m$ , the refractive index of the ice cube is

1

\frac{4}{3}

2

\frac{6}{5}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

supporting img
Apparent depth $= ($ real depth $) / μ$ .
From one side
$x / μ = 12 c m (1)$
From opposite side
$(24 - x) / μ = 4 c m (2)$
Divide (1) by (2), we get
$\frac{x / μ}{(24 - x) / μ} = \frac{12}{4} \Rightarrow \frac{x}{24 - x} = 3 \Rightarrow x = 18$
$μ = \frac{18}{12} = \frac{3}{2}$

JEE - 2023

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364899 A coin is present on the bottom of a water tank of depth $30 c m .$ If the refractive index of the water is 1.5 then shift in the position of the coin as seen by an observer standing in air is

1

10 c m

2

20 c m

3

5 c m

4

2 c m

Explanation:

Shift $= d (1 - \frac{1}{μ}) = 30 (1 - \frac{1}{1.5}) = 10 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364900 When the rectangular metal tank is filled up to the top with an unknown liquid, an observer with eyes level with the top of the tank can just see the corner $E$ . Find the index of refraction of the liquid
supporting img

1 1.5

2 1.25

3 1.67

4 2.4

Explanation:

In $△ A B C$
$Sin θ_{c} = \frac{60}{\sqrt{60^{2} + 80^{2}}} = \frac{60}{100}$

supporting img

Also, $\sin θ_{C} = \frac{1}{μ}$
$\Rightarrow \frac{60}{100} = \frac{1}{μ} [∵ Sin θ c = \frac{100}{60}] \Rightarrow μ = 1.67$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364901 A fish at a depth of $12 c m$ in water $(μ = \frac{3}{4})$ is viewed by an observer on the bank of a lake. Its apparent rise as observed by the observer

1

3 c m

2

9 c m

3

8 c m

4

4 c m

Explanation:

Apparent rise
$= d (1 - \frac{1}{μ}) = 12 \times (1 - \frac{3}{4}) = 3 c m$