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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364923 A transparent cube of $0.21 m$ edge contains a small air bubble. Its apparent distance when viewed through one face of the cube is $0.10 m$ and when viewed from the opposite face is $0.04 m$ . The actual distance of the bubble from the second face of the cube

1

0.06 m

2

0.17 m

3

0.05 m

4

0.04 m

Explanation:

supporting img
$μ = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e} = \frac{(0.21 - x)}{0.10} = \frac{x}{0.04}$
$\Rightarrow 0.21 \times 0.04 - x \times 0.04 = 0.10 x$
$\Rightarrow x = \frac{0.21 \times 0.04}{0.14} = x = 0.06 m$

KCET - 2006

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364924 Each quarter of a vessel of depth $H$ is filled with liquids of the refractive indices $n_{1}, n_{2}, n_{3} a n d n_{4}$ from the bottom respectively. The apparent depth of the vessel when looked normally is

1

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{4}

2

\frac{H (n_{1} + n_{2} + n_{3} + n_{4})}{4}

3

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{2}

4

\frac{(n_{1} + n_{2} + n_{3} + n_{4})}{4 H}

Explanation:

Apparent depth of bottom
$= \frac{H / 4}{n_{1}} + \frac{H / 4}{n_{2}} + \frac{H / 4}{n_{3}} + \frac{H / 4}{n_{4}}$
$= \frac{H}{4} (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364925 The image of point $P$ when viewed from top of the slabs will be
supporting img

1

1.5 c m

above

P

2

2.0 c m

above

P

3

1 c m

above

P

4

2.0 c m

below

P

Explanation:

The two slabs will shift the image a distance
$d = 2 (1 - \frac{1}{μ}) t = 2 (1 - \frac{1}{1.5}) (1.5) = 1.0 c m$
Therefore, final image will be $1 c m$ above point $P$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364926 An air bubble in glass slab $(μ = 1.5)$ when viewed from one side appears to be at $3 c m$ and from opposite side at $4 c m$ . The thickness of the slab is

1

7 c m

2

9.3 c m

3

10.5 c m

4

12.0 c m

Explanation:

Thickness $= μ (x + y) = 1.5 (3 + 4)$
$= 10.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364927 A coin is placed at the bottom of a tank which is filled by two liquid layers as shown. The apparent depth of coin from top of surface for normal viewing is
supporting img

1

6 c m

2

13 c m

3

5 c m

4

18 c m

Explanation:

$A = \frac{R_{1}}{μ_{1}} + \frac{R_{2}}{μ_{2}}$
$\begin{aligned} = \frac{12}{4 / 3} + \frac{6}{3 / 2} \\ = 9 + 4 = 13 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364923 A transparent cube of $0.21 m$ edge contains a small air bubble. Its apparent distance when viewed through one face of the cube is $0.10 m$ and when viewed from the opposite face is $0.04 m$ . The actual distance of the bubble from the second face of the cube

1

0.06 m

2

0.17 m

3

0.05 m

4

0.04 m

Explanation:

supporting img
$μ = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e} = \frac{(0.21 - x)}{0.10} = \frac{x}{0.04}$
$\Rightarrow 0.21 \times 0.04 - x \times 0.04 = 0.10 x$
$\Rightarrow x = \frac{0.21 \times 0.04}{0.14} = x = 0.06 m$

KCET - 2006

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364924 Each quarter of a vessel of depth $H$ is filled with liquids of the refractive indices $n_{1}, n_{2}, n_{3} a n d n_{4}$ from the bottom respectively. The apparent depth of the vessel when looked normally is

1

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{4}

2

\frac{H (n_{1} + n_{2} + n_{3} + n_{4})}{4}

3

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{2}

4

\frac{(n_{1} + n_{2} + n_{3} + n_{4})}{4 H}

Explanation:

Apparent depth of bottom
$= \frac{H / 4}{n_{1}} + \frac{H / 4}{n_{2}} + \frac{H / 4}{n_{3}} + \frac{H / 4}{n_{4}}$
$= \frac{H}{4} (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364925 The image of point $P$ when viewed from top of the slabs will be
supporting img

1

1.5 c m

above

P

2

2.0 c m

above

P

3

1 c m

above

P

4

2.0 c m

below

P

Explanation:

The two slabs will shift the image a distance
$d = 2 (1 - \frac{1}{μ}) t = 2 (1 - \frac{1}{1.5}) (1.5) = 1.0 c m$
Therefore, final image will be $1 c m$ above point $P$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364926 An air bubble in glass slab $(μ = 1.5)$ when viewed from one side appears to be at $3 c m$ and from opposite side at $4 c m$ . The thickness of the slab is

1

7 c m

2

9.3 c m

3

10.5 c m

4

12.0 c m

Explanation:

Thickness $= μ (x + y) = 1.5 (3 + 4)$
$= 10.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364927 A coin is placed at the bottom of a tank which is filled by two liquid layers as shown. The apparent depth of coin from top of surface for normal viewing is
supporting img

1

6 c m

2

13 c m

3

5 c m

4

18 c m

Explanation:

$A = \frac{R_{1}}{μ_{1}} + \frac{R_{2}}{μ_{2}}$
$\begin{aligned} = \frac{12}{4 / 3} + \frac{6}{3 / 2} \\ = 9 + 4 = 13 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364923 A transparent cube of $0.21 m$ edge contains a small air bubble. Its apparent distance when viewed through one face of the cube is $0.10 m$ and when viewed from the opposite face is $0.04 m$ . The actual distance of the bubble from the second face of the cube

1

0.06 m

2

0.17 m

3

0.05 m

4

0.04 m

Explanation:

supporting img
$μ = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e} = \frac{(0.21 - x)}{0.10} = \frac{x}{0.04}$
$\Rightarrow 0.21 \times 0.04 - x \times 0.04 = 0.10 x$
$\Rightarrow x = \frac{0.21 \times 0.04}{0.14} = x = 0.06 m$

KCET - 2006

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364924 Each quarter of a vessel of depth $H$ is filled with liquids of the refractive indices $n_{1}, n_{2}, n_{3} a n d n_{4}$ from the bottom respectively. The apparent depth of the vessel when looked normally is

1

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{4}

2

\frac{H (n_{1} + n_{2} + n_{3} + n_{4})}{4}

3

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{2}

4

\frac{(n_{1} + n_{2} + n_{3} + n_{4})}{4 H}

Explanation:

Apparent depth of bottom
$= \frac{H / 4}{n_{1}} + \frac{H / 4}{n_{2}} + \frac{H / 4}{n_{3}} + \frac{H / 4}{n_{4}}$
$= \frac{H}{4} (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364925 The image of point $P$ when viewed from top of the slabs will be
supporting img

1

1.5 c m

above

P

2

2.0 c m

above

P

3

1 c m

above

P

4

2.0 c m

below

P

Explanation:

The two slabs will shift the image a distance
$d = 2 (1 - \frac{1}{μ}) t = 2 (1 - \frac{1}{1.5}) (1.5) = 1.0 c m$
Therefore, final image will be $1 c m$ above point $P$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364926 An air bubble in glass slab $(μ = 1.5)$ when viewed from one side appears to be at $3 c m$ and from opposite side at $4 c m$ . The thickness of the slab is

1

7 c m

2

9.3 c m

3

10.5 c m

4

12.0 c m

Explanation:

Thickness $= μ (x + y) = 1.5 (3 + 4)$
$= 10.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364927 A coin is placed at the bottom of a tank which is filled by two liquid layers as shown. The apparent depth of coin from top of surface for normal viewing is
supporting img

1

6 c m

2

13 c m

3

5 c m

4

18 c m

Explanation:

$A = \frac{R_{1}}{μ_{1}} + \frac{R_{2}}{μ_{2}}$
$\begin{aligned} = \frac{12}{4 / 3} + \frac{6}{3 / 2} \\ = 9 + 4 = 13 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364923 A transparent cube of $0.21 m$ edge contains a small air bubble. Its apparent distance when viewed through one face of the cube is $0.10 m$ and when viewed from the opposite face is $0.04 m$ . The actual distance of the bubble from the second face of the cube

1

0.06 m

2

0.17 m

3

0.05 m

4

0.04 m

Explanation:

supporting img
$μ = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e} = \frac{(0.21 - x)}{0.10} = \frac{x}{0.04}$
$\Rightarrow 0.21 \times 0.04 - x \times 0.04 = 0.10 x$
$\Rightarrow x = \frac{0.21 \times 0.04}{0.14} = x = 0.06 m$

KCET - 2006

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364924 Each quarter of a vessel of depth $H$ is filled with liquids of the refractive indices $n_{1}, n_{2}, n_{3} a n d n_{4}$ from the bottom respectively. The apparent depth of the vessel when looked normally is

1

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{4}

2

\frac{H (n_{1} + n_{2} + n_{3} + n_{4})}{4}

3

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{2}

4

\frac{(n_{1} + n_{2} + n_{3} + n_{4})}{4 H}

Explanation:

Apparent depth of bottom
$= \frac{H / 4}{n_{1}} + \frac{H / 4}{n_{2}} + \frac{H / 4}{n_{3}} + \frac{H / 4}{n_{4}}$
$= \frac{H}{4} (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364925 The image of point $P$ when viewed from top of the slabs will be
supporting img

1

1.5 c m

above

P

2

2.0 c m

above

P

3

1 c m

above

P

4

2.0 c m

below

P

Explanation:

The two slabs will shift the image a distance
$d = 2 (1 - \frac{1}{μ}) t = 2 (1 - \frac{1}{1.5}) (1.5) = 1.0 c m$
Therefore, final image will be $1 c m$ above point $P$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364926 An air bubble in glass slab $(μ = 1.5)$ when viewed from one side appears to be at $3 c m$ and from opposite side at $4 c m$ . The thickness of the slab is

1

7 c m

2

9.3 c m

3

10.5 c m

4

12.0 c m

Explanation:

Thickness $= μ (x + y) = 1.5 (3 + 4)$
$= 10.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364927 A coin is placed at the bottom of a tank which is filled by two liquid layers as shown. The apparent depth of coin from top of surface for normal viewing is
supporting img

1

6 c m

2

13 c m

3

5 c m

4

18 c m

Explanation:

$A = \frac{R_{1}}{μ_{1}} + \frac{R_{2}}{μ_{2}}$
$\begin{aligned} = \frac{12}{4 / 3} + \frac{6}{3 / 2} \\ = 9 + 4 = 13 c m \end{aligned}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364923 A transparent cube of $0.21 m$ edge contains a small air bubble. Its apparent distance when viewed through one face of the cube is $0.10 m$ and when viewed from the opposite face is $0.04 m$ . The actual distance of the bubble from the second face of the cube

1

0.06 m

2

0.17 m

3

0.05 m

4

0.04 m

Explanation:

supporting img
$μ = \frac{r e a l d i s t a n c e}{a p p a r e n t d i s t a n c e} = \frac{(0.21 - x)}{0.10} = \frac{x}{0.04}$
$\Rightarrow 0.21 \times 0.04 - x \times 0.04 = 0.10 x$
$\Rightarrow x = \frac{0.21 \times 0.04}{0.14} = x = 0.06 m$

KCET - 2006

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364924 Each quarter of a vessel of depth $H$ is filled with liquids of the refractive indices $n_{1}, n_{2}, n_{3} a n d n_{4}$ from the bottom respectively. The apparent depth of the vessel when looked normally is

1

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{4}

2

\frac{H (n_{1} + n_{2} + n_{3} + n_{4})}{4}

3

\frac{H (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})}{2}

4

\frac{(n_{1} + n_{2} + n_{3} + n_{4})}{4 H}

Explanation:

Apparent depth of bottom
$= \frac{H / 4}{n_{1}} + \frac{H / 4}{n_{2}} + \frac{H / 4}{n_{3}} + \frac{H / 4}{n_{4}}$
$= \frac{H}{4} (\frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}} + \frac{1}{n_{4}})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364925 The image of point $P$ when viewed from top of the slabs will be
supporting img

1

1.5 c m

above

P

2

2.0 c m

above

P

3

1 c m

above

P

4

2.0 c m

below

P

Explanation:

The two slabs will shift the image a distance
$d = 2 (1 - \frac{1}{μ}) t = 2 (1 - \frac{1}{1.5}) (1.5) = 1.0 c m$
Therefore, final image will be $1 c m$ above point $P$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364926 An air bubble in glass slab $(μ = 1.5)$ when viewed from one side appears to be at $3 c m$ and from opposite side at $4 c m$ . The thickness of the slab is

1

7 c m

2

9.3 c m

3

10.5 c m

4

12.0 c m

Explanation:

Thickness $= μ (x + y) = 1.5 (3 + 4)$
$= 10.5 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364927 A coin is placed at the bottom of a tank which is filled by two liquid layers as shown. The apparent depth of coin from top of surface for normal viewing is
supporting img

1

6 c m

2

13 c m

3

5 c m

4

18 c m

Explanation:

$A = \frac{R_{1}}{μ_{1}} + \frac{R_{2}}{μ_{2}}$
$\begin{aligned} = \frac{12}{4 / 3} + \frac{6}{3 / 2} \\ = 9 + 4 = 13 c m \end{aligned}$