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Ray Optics

282217 The apparent depth of a needle lying in a water beaker is found to be $9 cm$ . If water is replaced by a liquid of refractive index 1.5 , then the apparent depth of needle will be ( $μ$ of water is 4/3)

1

10 cm

2

9 cm

3

12 cm

4

7 cm

(e)

8 cm

Explanation:

E : Given that, $x$ is the real depth
Apparent depth, $y = 9 cm$
Refractive index of liquid $(μ_{l}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know,
$μ_{w} = \frac{Realdepth (x)}{Apparent depth (y)}$
$\begin{array}{r} \frac{4}{3} = \frac{x}{9} \\ x = 12 cm \\ μ_{l} = \frac{Real depth (x)}{Apparent depth (y^{'})} \\ y^{'} = 12 \times \frac{1}{1.5} = 8 cm \end{array}$

Kerala CEE 2021

Ray Optics

282218 If refractive index of water is $\frac{4}{3}$ and that of a given slab immersed in it is $\frac{5}{3}$ . The value of critical angle for a ray of light tending to go from glass to water is :

1

{Sin}^{- 1} (\frac{4}{5})

2

{Sin}^{- 1} (\frac{3}{4})

3

{Sin}^{- 1} (\frac{5}{3})

4

{Sin}^{- 1} (\frac{4}{3})

Explanation:

A: Given, $μ_{w} = 4 / 3, μ_{s} = 5 / 3$
$_{w} μ_{s} = \frac{μ_{s}}{μ_{w}} = \frac{5 / 3}{4 / 3} = \frac{5}{4}$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$θ_{c} = \sin^{- 1} (\frac{4}{5})$

AP EAMCET-25.08.2021

Ray Optics

282219 A well-cut diamond appears bright because

1 it emits light

2 it is radioactive

3 of its total internal reflection

4 it has high density

AP EAMCET-23.08.2021

Ray Optics

282220 Conditions for total internal reflection to occur are

1

(b, c)

2

(a, c)

3

(a, d)

4

(b, d)

TS EAMCET (Medical) 09.08.2021

Ray Optics

282217 The apparent depth of a needle lying in a water beaker is found to be $9 cm$ . If water is replaced by a liquid of refractive index 1.5 , then the apparent depth of needle will be ( $μ$ of water is 4/3)

1

10 cm

2

9 cm

3

12 cm

4

7 cm

(e)

8 cm

Explanation:

E : Given that, $x$ is the real depth
Apparent depth, $y = 9 cm$
Refractive index of liquid $(μ_{l}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know,
$μ_{w} = \frac{Realdepth (x)}{Apparent depth (y)}$
$\begin{array}{r} \frac{4}{3} = \frac{x}{9} \\ x = 12 cm \\ μ_{l} = \frac{Real depth (x)}{Apparent depth (y^{'})} \\ y^{'} = 12 \times \frac{1}{1.5} = 8 cm \end{array}$

Kerala CEE 2021

Ray Optics

282218 If refractive index of water is $\frac{4}{3}$ and that of a given slab immersed in it is $\frac{5}{3}$ . The value of critical angle for a ray of light tending to go from glass to water is :

1

{Sin}^{- 1} (\frac{4}{5})

2

{Sin}^{- 1} (\frac{3}{4})

3

{Sin}^{- 1} (\frac{5}{3})

4

{Sin}^{- 1} (\frac{4}{3})

Explanation:

A: Given, $μ_{w} = 4 / 3, μ_{s} = 5 / 3$
$_{w} μ_{s} = \frac{μ_{s}}{μ_{w}} = \frac{5 / 3}{4 / 3} = \frac{5}{4}$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$θ_{c} = \sin^{- 1} (\frac{4}{5})$

AP EAMCET-25.08.2021

Ray Optics

282219 A well-cut diamond appears bright because

1 it emits light

2 it is radioactive

3 of its total internal reflection

4 it has high density

AP EAMCET-23.08.2021

Ray Optics

282220 Conditions for total internal reflection to occur are

1

(b, c)

2

(a, c)

3

(a, d)

4

(b, d)

TS EAMCET (Medical) 09.08.2021

Ray Optics

282217 The apparent depth of a needle lying in a water beaker is found to be $9 cm$ . If water is replaced by a liquid of refractive index 1.5 , then the apparent depth of needle will be ( $μ$ of water is 4/3)

1

10 cm

2

9 cm

3

12 cm

4

7 cm

(e)

8 cm

Explanation:

E : Given that, $x$ is the real depth
Apparent depth, $y = 9 cm$
Refractive index of liquid $(μ_{l}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know,
$μ_{w} = \frac{Realdepth (x)}{Apparent depth (y)}$
$\begin{array}{r} \frac{4}{3} = \frac{x}{9} \\ x = 12 cm \\ μ_{l} = \frac{Real depth (x)}{Apparent depth (y^{'})} \\ y^{'} = 12 \times \frac{1}{1.5} = 8 cm \end{array}$

Kerala CEE 2021

Ray Optics

282218 If refractive index of water is $\frac{4}{3}$ and that of a given slab immersed in it is $\frac{5}{3}$ . The value of critical angle for a ray of light tending to go from glass to water is :

1

{Sin}^{- 1} (\frac{4}{5})

2

{Sin}^{- 1} (\frac{3}{4})

3

{Sin}^{- 1} (\frac{5}{3})

4

{Sin}^{- 1} (\frac{4}{3})

Explanation:

A: Given, $μ_{w} = 4 / 3, μ_{s} = 5 / 3$
$_{w} μ_{s} = \frac{μ_{s}}{μ_{w}} = \frac{5 / 3}{4 / 3} = \frac{5}{4}$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$θ_{c} = \sin^{- 1} (\frac{4}{5})$

AP EAMCET-25.08.2021

Ray Optics

282219 A well-cut diamond appears bright because

1 it emits light

2 it is radioactive

3 of its total internal reflection

4 it has high density

AP EAMCET-23.08.2021

Ray Optics

282220 Conditions for total internal reflection to occur are

1

(b, c)

2

(a, c)

3

(a, d)

4

(b, d)

TS EAMCET (Medical) 09.08.2021

Ray Optics

282217 The apparent depth of a needle lying in a water beaker is found to be $9 cm$ . If water is replaced by a liquid of refractive index 1.5 , then the apparent depth of needle will be ( $μ$ of water is 4/3)

1

10 cm

2

9 cm

3

12 cm

4

7 cm

(e)

8 cm

Explanation:

E : Given that, $x$ is the real depth
Apparent depth, $y = 9 cm$
Refractive index of liquid $(μ_{l}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know,
$μ_{w} = \frac{Realdepth (x)}{Apparent depth (y)}$
$\begin{array}{r} \frac{4}{3} = \frac{x}{9} \\ x = 12 cm \\ μ_{l} = \frac{Real depth (x)}{Apparent depth (y^{'})} \\ y^{'} = 12 \times \frac{1}{1.5} = 8 cm \end{array}$

Kerala CEE 2021

Ray Optics

282218 If refractive index of water is $\frac{4}{3}$ and that of a given slab immersed in it is $\frac{5}{3}$ . The value of critical angle for a ray of light tending to go from glass to water is :

1

{Sin}^{- 1} (\frac{4}{5})

2

{Sin}^{- 1} (\frac{3}{4})

3

{Sin}^{- 1} (\frac{5}{3})

4

{Sin}^{- 1} (\frac{4}{3})

Explanation:

A: Given, $μ_{w} = 4 / 3, μ_{s} = 5 / 3$
$_{w} μ_{s} = \frac{μ_{s}}{μ_{w}} = \frac{5 / 3}{4 / 3} = \frac{5}{4}$
Critical angle $(θ_{c}) = \sin^{- 1} (\frac{1}{μ})$
$θ_{c} = \sin^{- 1} (\frac{4}{5})$

AP EAMCET-25.08.2021

Ray Optics

282219 A well-cut diamond appears bright because

1 it emits light

2 it is radioactive

3 of its total internal reflection

4 it has high density

AP EAMCET-23.08.2021

Ray Optics

282220 Conditions for total internal reflection to occur are

1

(b, c)

2

(a, c)

3

(a, d)

4

(b, d)

TS EAMCET (Medical) 09.08.2021

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