QBManager

Ray Optics

282097 Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2 n}$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be

1

\sin^{- 1} (\sqrt{n})

2

\cos^{- 1} (\sqrt{\frac{n}{2}})

3

\sin^{- 1} (\sqrt{2 n})

4

2 \cos^{- 1} (\sqrt{\frac{n}{2}})

Explanation:

Given that, $r = \frac{i}{2}$ and refractive index $(μ) = \sqrt{2 n}$ Refractive index,
$\begin{array}{r} μ = \frac{\sin i}{\sin r} \\ \sqrt{2 n} = \frac{\sin i}{\sin (\frac{i}{2})} = \frac{2 \sin (\frac{i}{2}) \cdot \cos (\frac{i}{2})}{\sin (\frac{i}{2})} \\ \sqrt{2 n} = 2 \cos (\frac{i}{2}) \\ \cos (\frac{i}{2}) = \sqrt{\frac{n}{2}} \\ i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}}) \end{array}$

JEE Main-27.06.2022

Ray Optics

282098 Time taken by light to travel in two different materials $A$ and $B$ of refractive indices is $μ_{A}$ and $μ_{B}$ of same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2} - t_{1} = 5 \times 10^{- 10} s$ and the ratio of $μ_{A}$ to $μ_{B}$ is $1 : 2$ . then the thickness of material, in meter is: (Given $v_{A}$ and $v_{B}$ are velocities of light in $A$ and $B$ materials respectively).

1

5 \times 10^{- 10} {VAm}_{A}

2

5 \times 10^{- 10} v_{B} m

3

1.5 \times 10^{- 10} v_{A} m

4

5 \times 10^{- 10} v_{B} m

Explanation:

A: Given that,
$\begin{array}{r} (t_{2} - t_{1}) = 5 \times 10^{- 10} \sec \\ \frac{μ_{A}}{μ_{B}} = \frac{1}{2} \\ \frac{μ_{A}}{μ_{B}} = \frac{c / v_{A}}{c / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2} \end{array}$
Let, thickness is 'y'
Then,
$\begin{array}{r} \frac{y}{v_{B}} - \frac{y}{v_{A}} = 5 \times 10^{- 10} \\ y = \frac{5 \times 10^{- 10} \times v_{A} v_{B}}{v_{A} - v_{B}} \end{array}$
\{from given\}
As $v_{A} = 2 v_{B}$
Hence, $y = \frac{5 \times 10^{- 10} \times 2 v_{B} \times v_{B}}{2 v_{B} - v_{B}}$
$y = 5 \times 10^{- 10} \times 2 v_{B}$
o1
$y = 5 \times 10^{- 10} \times v_{A}$
${∵ v_{A} = 2 v_{B}}$

JEE Main-25.07.2022

Ray Optics

282099 The difference of speed of light in the two media $A$ and $B (v_{A} - v_{B})$ is $2.6 \times 10^{7} m / s$ . If the refractive index of medium $B$ is 1.47 , then the ratio of refractive index of medium $B$ to medium $A$ is : (Given : speed of light in vacuum $c = 3 \times 10^{8} {ms}^{- 1}$ )

1 1.303

2 1.318

3 1.13

4 0.12

Explanation:

C: Given that,
Difference of speed of light $(v_{A} - v_{B}) = 2.6 \times 10^{7} m / s$
Refractive index of medium $B (μ_{B}) = 1.47$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$\begin{aligned} μ_{B} & = \frac{c}{v_{B}} \\ v_{B} & = \frac{c}{μ_{B}} \\ v_{B} & = \frac{3 \times 10^{8}}{1.47} = 2.04 \times 10^{8} m / s \end{aligned}$
Now, $v_{A} = (2.6 \times 10^{7} + v_{B}) m / s$
$\begin{aligned} A & = (2.6 \times 26 \times 10^{8} + 2.04 \times 10^{8}) m / s \\ v_{A} & = (0.26 \times 10^{8} m / s \\ v_{A} & = 2.3 \times 10^{2} \end{aligned}$
Again, $μ_{A} = \frac{c}{v_{A}}$
$\begin{aligned} μ_{A} & = \frac{3 \times 10^{8}}{2.3 \times 10^{8}} \\ μ_{A} & = 1.3 \end{aligned}$
Hence, $\frac{μ_{B}}{μ_{A}} = \frac{1.47}{1.3} = 1.13$

JEE Main-25.06.2022

Ray Optics

282100 Which of the following statement is correct?

1 In primary rainbow, observer sees red colour on the top and violet on the bottom

2 In primary rainbow, observer sees violet colour on the top and red on the bottom

3 In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops

4 Primary rainbow is less bright than secondary rainbow,

JEE Main-25.07.2022

Ray Optics

282097 Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2 n}$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be

1

\sin^{- 1} (\sqrt{n})

2

\cos^{- 1} (\sqrt{\frac{n}{2}})

3

\sin^{- 1} (\sqrt{2 n})

4

2 \cos^{- 1} (\sqrt{\frac{n}{2}})

Explanation:

Given that, $r = \frac{i}{2}$ and refractive index $(μ) = \sqrt{2 n}$ Refractive index,
$\begin{array}{r} μ = \frac{\sin i}{\sin r} \\ \sqrt{2 n} = \frac{\sin i}{\sin (\frac{i}{2})} = \frac{2 \sin (\frac{i}{2}) \cdot \cos (\frac{i}{2})}{\sin (\frac{i}{2})} \\ \sqrt{2 n} = 2 \cos (\frac{i}{2}) \\ \cos (\frac{i}{2}) = \sqrt{\frac{n}{2}} \\ i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}}) \end{array}$

JEE Main-27.06.2022

Ray Optics

282098 Time taken by light to travel in two different materials $A$ and $B$ of refractive indices is $μ_{A}$ and $μ_{B}$ of same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2} - t_{1} = 5 \times 10^{- 10} s$ and the ratio of $μ_{A}$ to $μ_{B}$ is $1 : 2$ . then the thickness of material, in meter is: (Given $v_{A}$ and $v_{B}$ are velocities of light in $A$ and $B$ materials respectively).

1

5 \times 10^{- 10} {VAm}_{A}

2

5 \times 10^{- 10} v_{B} m

3

1.5 \times 10^{- 10} v_{A} m

4

5 \times 10^{- 10} v_{B} m

Explanation:

A: Given that,
$\begin{array}{r} (t_{2} - t_{1}) = 5 \times 10^{- 10} \sec \\ \frac{μ_{A}}{μ_{B}} = \frac{1}{2} \\ \frac{μ_{A}}{μ_{B}} = \frac{c / v_{A}}{c / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2} \end{array}$
Let, thickness is 'y'
Then,
$\begin{array}{r} \frac{y}{v_{B}} - \frac{y}{v_{A}} = 5 \times 10^{- 10} \\ y = \frac{5 \times 10^{- 10} \times v_{A} v_{B}}{v_{A} - v_{B}} \end{array}$
\{from given\}
As $v_{A} = 2 v_{B}$
Hence, $y = \frac{5 \times 10^{- 10} \times 2 v_{B} \times v_{B}}{2 v_{B} - v_{B}}$
$y = 5 \times 10^{- 10} \times 2 v_{B}$
o1
$y = 5 \times 10^{- 10} \times v_{A}$
${∵ v_{A} = 2 v_{B}}$

JEE Main-25.07.2022

Ray Optics

282099 The difference of speed of light in the two media $A$ and $B (v_{A} - v_{B})$ is $2.6 \times 10^{7} m / s$ . If the refractive index of medium $B$ is 1.47 , then the ratio of refractive index of medium $B$ to medium $A$ is : (Given : speed of light in vacuum $c = 3 \times 10^{8} {ms}^{- 1}$ )

1 1.303

2 1.318

3 1.13

4 0.12

Explanation:

C: Given that,
Difference of speed of light $(v_{A} - v_{B}) = 2.6 \times 10^{7} m / s$
Refractive index of medium $B (μ_{B}) = 1.47$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$\begin{aligned} μ_{B} & = \frac{c}{v_{B}} \\ v_{B} & = \frac{c}{μ_{B}} \\ v_{B} & = \frac{3 \times 10^{8}}{1.47} = 2.04 \times 10^{8} m / s \end{aligned}$
Now, $v_{A} = (2.6 \times 10^{7} + v_{B}) m / s$
$\begin{aligned} A & = (2.6 \times 26 \times 10^{8} + 2.04 \times 10^{8}) m / s \\ v_{A} & = (0.26 \times 10^{8} m / s \\ v_{A} & = 2.3 \times 10^{2} \end{aligned}$
Again, $μ_{A} = \frac{c}{v_{A}}$
$\begin{aligned} μ_{A} & = \frac{3 \times 10^{8}}{2.3 \times 10^{8}} \\ μ_{A} & = 1.3 \end{aligned}$
Hence, $\frac{μ_{B}}{μ_{A}} = \frac{1.47}{1.3} = 1.13$

JEE Main-25.06.2022

Ray Optics

282100 Which of the following statement is correct?

1 In primary rainbow, observer sees red colour on the top and violet on the bottom

2 In primary rainbow, observer sees violet colour on the top and red on the bottom

3 In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops

4 Primary rainbow is less bright than secondary rainbow,

JEE Main-25.07.2022

Ray Optics

282097 Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2 n}$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be

1

\sin^{- 1} (\sqrt{n})

2

\cos^{- 1} (\sqrt{\frac{n}{2}})

3

\sin^{- 1} (\sqrt{2 n})

4

2 \cos^{- 1} (\sqrt{\frac{n}{2}})

Explanation:

Given that, $r = \frac{i}{2}$ and refractive index $(μ) = \sqrt{2 n}$ Refractive index,
$\begin{array}{r} μ = \frac{\sin i}{\sin r} \\ \sqrt{2 n} = \frac{\sin i}{\sin (\frac{i}{2})} = \frac{2 \sin (\frac{i}{2}) \cdot \cos (\frac{i}{2})}{\sin (\frac{i}{2})} \\ \sqrt{2 n} = 2 \cos (\frac{i}{2}) \\ \cos (\frac{i}{2}) = \sqrt{\frac{n}{2}} \\ i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}}) \end{array}$

JEE Main-27.06.2022

Ray Optics

282098 Time taken by light to travel in two different materials $A$ and $B$ of refractive indices is $μ_{A}$ and $μ_{B}$ of same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2} - t_{1} = 5 \times 10^{- 10} s$ and the ratio of $μ_{A}$ to $μ_{B}$ is $1 : 2$ . then the thickness of material, in meter is: (Given $v_{A}$ and $v_{B}$ are velocities of light in $A$ and $B$ materials respectively).

1

5 \times 10^{- 10} {VAm}_{A}

2

5 \times 10^{- 10} v_{B} m

3

1.5 \times 10^{- 10} v_{A} m

4

5 \times 10^{- 10} v_{B} m

Explanation:

A: Given that,
$\begin{array}{r} (t_{2} - t_{1}) = 5 \times 10^{- 10} \sec \\ \frac{μ_{A}}{μ_{B}} = \frac{1}{2} \\ \frac{μ_{A}}{μ_{B}} = \frac{c / v_{A}}{c / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2} \end{array}$
Let, thickness is 'y'
Then,
$\begin{array}{r} \frac{y}{v_{B}} - \frac{y}{v_{A}} = 5 \times 10^{- 10} \\ y = \frac{5 \times 10^{- 10} \times v_{A} v_{B}}{v_{A} - v_{B}} \end{array}$
\{from given\}
As $v_{A} = 2 v_{B}$
Hence, $y = \frac{5 \times 10^{- 10} \times 2 v_{B} \times v_{B}}{2 v_{B} - v_{B}}$
$y = 5 \times 10^{- 10} \times 2 v_{B}$
o1
$y = 5 \times 10^{- 10} \times v_{A}$
${∵ v_{A} = 2 v_{B}}$

JEE Main-25.07.2022

Ray Optics

282099 The difference of speed of light in the two media $A$ and $B (v_{A} - v_{B})$ is $2.6 \times 10^{7} m / s$ . If the refractive index of medium $B$ is 1.47 , then the ratio of refractive index of medium $B$ to medium $A$ is : (Given : speed of light in vacuum $c = 3 \times 10^{8} {ms}^{- 1}$ )

1 1.303

2 1.318

3 1.13

4 0.12

Explanation:

C: Given that,
Difference of speed of light $(v_{A} - v_{B}) = 2.6 \times 10^{7} m / s$
Refractive index of medium $B (μ_{B}) = 1.47$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$\begin{aligned} μ_{B} & = \frac{c}{v_{B}} \\ v_{B} & = \frac{c}{μ_{B}} \\ v_{B} & = \frac{3 \times 10^{8}}{1.47} = 2.04 \times 10^{8} m / s \end{aligned}$
Now, $v_{A} = (2.6 \times 10^{7} + v_{B}) m / s$
$\begin{aligned} A & = (2.6 \times 26 \times 10^{8} + 2.04 \times 10^{8}) m / s \\ v_{A} & = (0.26 \times 10^{8} m / s \\ v_{A} & = 2.3 \times 10^{2} \end{aligned}$
Again, $μ_{A} = \frac{c}{v_{A}}$
$\begin{aligned} μ_{A} & = \frac{3 \times 10^{8}}{2.3 \times 10^{8}} \\ μ_{A} & = 1.3 \end{aligned}$
Hence, $\frac{μ_{B}}{μ_{A}} = \frac{1.47}{1.3} = 1.13$

JEE Main-25.06.2022

Ray Optics

282100 Which of the following statement is correct?

1 In primary rainbow, observer sees red colour on the top and violet on the bottom

2 In primary rainbow, observer sees violet colour on the top and red on the bottom

3 In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops

4 Primary rainbow is less bright than secondary rainbow,

JEE Main-25.07.2022

Ray Optics

282097 Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2 n}$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be

1

\sin^{- 1} (\sqrt{n})

2

\cos^{- 1} (\sqrt{\frac{n}{2}})

3

\sin^{- 1} (\sqrt{2 n})

4

2 \cos^{- 1} (\sqrt{\frac{n}{2}})

Explanation:

Given that, $r = \frac{i}{2}$ and refractive index $(μ) = \sqrt{2 n}$ Refractive index,
$\begin{array}{r} μ = \frac{\sin i}{\sin r} \\ \sqrt{2 n} = \frac{\sin i}{\sin (\frac{i}{2})} = \frac{2 \sin (\frac{i}{2}) \cdot \cos (\frac{i}{2})}{\sin (\frac{i}{2})} \\ \sqrt{2 n} = 2 \cos (\frac{i}{2}) \\ \cos (\frac{i}{2}) = \sqrt{\frac{n}{2}} \\ i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}}) \end{array}$

JEE Main-27.06.2022

Ray Optics

282098 Time taken by light to travel in two different materials $A$ and $B$ of refractive indices is $μ_{A}$ and $μ_{B}$ of same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2} - t_{1} = 5 \times 10^{- 10} s$ and the ratio of $μ_{A}$ to $μ_{B}$ is $1 : 2$ . then the thickness of material, in meter is: (Given $v_{A}$ and $v_{B}$ are velocities of light in $A$ and $B$ materials respectively).

1

5 \times 10^{- 10} {VAm}_{A}

2

5 \times 10^{- 10} v_{B} m

3

1.5 \times 10^{- 10} v_{A} m

4

5 \times 10^{- 10} v_{B} m

Explanation:

A: Given that,
$\begin{array}{r} (t_{2} - t_{1}) = 5 \times 10^{- 10} \sec \\ \frac{μ_{A}}{μ_{B}} = \frac{1}{2} \\ \frac{μ_{A}}{μ_{B}} = \frac{c / v_{A}}{c / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2} \end{array}$
Let, thickness is 'y'
Then,
$\begin{array}{r} \frac{y}{v_{B}} - \frac{y}{v_{A}} = 5 \times 10^{- 10} \\ y = \frac{5 \times 10^{- 10} \times v_{A} v_{B}}{v_{A} - v_{B}} \end{array}$
\{from given\}
As $v_{A} = 2 v_{B}$
Hence, $y = \frac{5 \times 10^{- 10} \times 2 v_{B} \times v_{B}}{2 v_{B} - v_{B}}$
$y = 5 \times 10^{- 10} \times 2 v_{B}$
o1
$y = 5 \times 10^{- 10} \times v_{A}$
${∵ v_{A} = 2 v_{B}}$

JEE Main-25.07.2022

Ray Optics

282099 The difference of speed of light in the two media $A$ and $B (v_{A} - v_{B})$ is $2.6 \times 10^{7} m / s$ . If the refractive index of medium $B$ is 1.47 , then the ratio of refractive index of medium $B$ to medium $A$ is : (Given : speed of light in vacuum $c = 3 \times 10^{8} {ms}^{- 1}$ )

1 1.303

2 1.318

3 1.13

4 0.12

Explanation:

C: Given that,
Difference of speed of light $(v_{A} - v_{B}) = 2.6 \times 10^{7} m / s$
Refractive index of medium $B (μ_{B}) = 1.47$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$\begin{aligned} μ_{B} & = \frac{c}{v_{B}} \\ v_{B} & = \frac{c}{μ_{B}} \\ v_{B} & = \frac{3 \times 10^{8}}{1.47} = 2.04 \times 10^{8} m / s \end{aligned}$
Now, $v_{A} = (2.6 \times 10^{7} + v_{B}) m / s$
$\begin{aligned} A & = (2.6 \times 26 \times 10^{8} + 2.04 \times 10^{8}) m / s \\ v_{A} & = (0.26 \times 10^{8} m / s \\ v_{A} & = 2.3 \times 10^{2} \end{aligned}$
Again, $μ_{A} = \frac{c}{v_{A}}$
$\begin{aligned} μ_{A} & = \frac{3 \times 10^{8}}{2.3 \times 10^{8}} \\ μ_{A} & = 1.3 \end{aligned}$
Hence, $\frac{μ_{B}}{μ_{A}} = \frac{1.47}{1.3} = 1.13$

JEE Main-25.06.2022

Ray Optics

282100 Which of the following statement is correct?

1 In primary rainbow, observer sees red colour on the top and violet on the bottom

2 In primary rainbow, observer sees violet colour on the top and red on the bottom

3 In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops

4 Primary rainbow is less bright than secondary rainbow,

JEE Main-25.07.2022

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