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Ray Optics

282351 When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:

1 dispersion of light

2 reflection of light

3 polarization of light

4 interference of light

Assam CEE-2017

Ray Optics

282352 The plane face of a Plano convex lens is silvered. If $μ$ be the refractive index and $R$ is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature

1

μ R

2

R^{2} / μ

3

R / (μ - 1)

4

[(μ + 1) / (μ - 1)] R

Explanation:

C: Consider a parallel beam of light incident on convex surface,
$\begin{array}{r} \frac{μ}{v_{1}} - \frac{1}{- \infty} = \frac{μ - 1}{R} \\ v_{1} = \frac{μ R}{μ - 1} \end{array}$
This image gets reflected from the mirror to form an image at distance ${\frac{μ R}{μ - 1}}$ to the left of lens.
Refraction again from convex surface-
$\begin{array}{r} \frac{1}{V_{2}} - \frac{μ}{\frac{μ R}{μ - 1}} = \frac{1 - μ}{- R} \\ \frac{1}{V_{2}} = \frac{2 (μ - 1)}{R} \\ \frac{1}{f} = \frac{2 (μ - 1)}{R} \\ f = \frac{R}{2 (μ - 1)} \\ R^{'} = 2 f = \frac{R}{μ - 1} \end{array}$
(left of side)
Hence, $R^{'} = \frac{R}{μ - 1}$

JIPMER-2017

Ray Optics

282353 In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30 cm$ . The focal length of the objective, when the angular magnification of the telescope is 2 , will be:

1

20 cm

2

30 cm

3

10 cm

4

15 cm

Explanation:

AGiven, $f_{0} + f_{e} = 30 cm$
And,
$\begin{array}{r} m = \frac{f_{0}}{f_{e}} = 2 \\ f_{0} = 2 f_{e} \end{array}$
Putting the value $f_{0}$ in equation (i), we get-
$\begin{array}{r} 2 f_{e} + f_{e} = 30 \\ f_{e} = \frac{30}{3} = 10 cm \end{array}$
Then,
$f_{0} = 2 f_{c} = 2 \times 10$
So, $f_{0} = 20 cm$

JEE Main-28.07.2022

Ray Optics

282354 The power of a lens (biconvex) is $1.25 m^{- 1}$ in particular medium. Refractive index of the lens is $1.5$ and radii of curvature are $20 cm$ and 40 cm respectively. The refractive index of surrounding medium:

1 1.0

2

\frac{9}{7}

3

\frac{3}{2}

4

\frac{4}{3}

Explanation:

B: Given, Power $(P) = 1.25 m^{- 1}$
Refractive index of lens $(μ_{g}) = 1.5$
Radius of curvature are, $R_{1} = 20 cm & R_{2} = - 40 cm$ Using lens maker formula-
$\begin{array}{r} P = \frac{1}{f} = (\frac{μ_{g}}{μ_{m}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{1}{0.2} + \frac{1}{0.4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{10}{2} + \frac{10}{4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (5 + \frac{5}{2}) \\ 1.25 = (\frac{1.5}{μ_{m}} - 1) (\frac{15}{2}) \\ \frac{2 \times 125}{15 \times 100} = (\frac{1.5}{μ_{m}} - 1) \\ \frac{2 \times 5}{15 \times 4} + 1 = \frac{3}{2 μ_{m}} \\ \frac{1}{3 \times 2} + 1 = \frac{3}{2 μ_{m}} \\ \frac{7}{6} = \frac{3}{2 μ} \\ μ_{m} = \frac{9}{7} \\ x \end{array}$

JEE Main-28.07.2022

Ray Optics

282351 When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:

1 dispersion of light

2 reflection of light

3 polarization of light

4 interference of light

Assam CEE-2017

Ray Optics

282352 The plane face of a Plano convex lens is silvered. If $μ$ be the refractive index and $R$ is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature

1

μ R

2

R^{2} / μ

3

R / (μ - 1)

4

[(μ + 1) / (μ - 1)] R

Explanation:

C: Consider a parallel beam of light incident on convex surface,
$\begin{array}{r} \frac{μ}{v_{1}} - \frac{1}{- \infty} = \frac{μ - 1}{R} \\ v_{1} = \frac{μ R}{μ - 1} \end{array}$
This image gets reflected from the mirror to form an image at distance ${\frac{μ R}{μ - 1}}$ to the left of lens.
Refraction again from convex surface-
$\begin{array}{r} \frac{1}{V_{2}} - \frac{μ}{\frac{μ R}{μ - 1}} = \frac{1 - μ}{- R} \\ \frac{1}{V_{2}} = \frac{2 (μ - 1)}{R} \\ \frac{1}{f} = \frac{2 (μ - 1)}{R} \\ f = \frac{R}{2 (μ - 1)} \\ R^{'} = 2 f = \frac{R}{μ - 1} \end{array}$
(left of side)
Hence, $R^{'} = \frac{R}{μ - 1}$

JIPMER-2017

Ray Optics

282353 In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30 cm$ . The focal length of the objective, when the angular magnification of the telescope is 2 , will be:

1

20 cm

2

30 cm

3

10 cm

4

15 cm

Explanation:

AGiven, $f_{0} + f_{e} = 30 cm$
And,
$\begin{array}{r} m = \frac{f_{0}}{f_{e}} = 2 \\ f_{0} = 2 f_{e} \end{array}$
Putting the value $f_{0}$ in equation (i), we get-
$\begin{array}{r} 2 f_{e} + f_{e} = 30 \\ f_{e} = \frac{30}{3} = 10 cm \end{array}$
Then,
$f_{0} = 2 f_{c} = 2 \times 10$
So, $f_{0} = 20 cm$

JEE Main-28.07.2022

Ray Optics

282354 The power of a lens (biconvex) is $1.25 m^{- 1}$ in particular medium. Refractive index of the lens is $1.5$ and radii of curvature are $20 cm$ and 40 cm respectively. The refractive index of surrounding medium:

1 1.0

2

\frac{9}{7}

3

\frac{3}{2}

4

\frac{4}{3}

Explanation:

B: Given, Power $(P) = 1.25 m^{- 1}$
Refractive index of lens $(μ_{g}) = 1.5$
Radius of curvature are, $R_{1} = 20 cm & R_{2} = - 40 cm$ Using lens maker formula-
$\begin{array}{r} P = \frac{1}{f} = (\frac{μ_{g}}{μ_{m}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{1}{0.2} + \frac{1}{0.4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{10}{2} + \frac{10}{4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (5 + \frac{5}{2}) \\ 1.25 = (\frac{1.5}{μ_{m}} - 1) (\frac{15}{2}) \\ \frac{2 \times 125}{15 \times 100} = (\frac{1.5}{μ_{m}} - 1) \\ \frac{2 \times 5}{15 \times 4} + 1 = \frac{3}{2 μ_{m}} \\ \frac{1}{3 \times 2} + 1 = \frac{3}{2 μ_{m}} \\ \frac{7}{6} = \frac{3}{2 μ} \\ μ_{m} = \frac{9}{7} \\ x \end{array}$

JEE Main-28.07.2022

Ray Optics

282351 When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:

1 dispersion of light

2 reflection of light

3 polarization of light

4 interference of light

Assam CEE-2017

Ray Optics

282352 The plane face of a Plano convex lens is silvered. If $μ$ be the refractive index and $R$ is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature

1

μ R

2

R^{2} / μ

3

R / (μ - 1)

4

[(μ + 1) / (μ - 1)] R

Explanation:

C: Consider a parallel beam of light incident on convex surface,
$\begin{array}{r} \frac{μ}{v_{1}} - \frac{1}{- \infty} = \frac{μ - 1}{R} \\ v_{1} = \frac{μ R}{μ - 1} \end{array}$
This image gets reflected from the mirror to form an image at distance ${\frac{μ R}{μ - 1}}$ to the left of lens.
Refraction again from convex surface-
$\begin{array}{r} \frac{1}{V_{2}} - \frac{μ}{\frac{μ R}{μ - 1}} = \frac{1 - μ}{- R} \\ \frac{1}{V_{2}} = \frac{2 (μ - 1)}{R} \\ \frac{1}{f} = \frac{2 (μ - 1)}{R} \\ f = \frac{R}{2 (μ - 1)} \\ R^{'} = 2 f = \frac{R}{μ - 1} \end{array}$
(left of side)
Hence, $R^{'} = \frac{R}{μ - 1}$

JIPMER-2017

Ray Optics

282353 In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30 cm$ . The focal length of the objective, when the angular magnification of the telescope is 2 , will be:

1

20 cm

2

30 cm

3

10 cm

4

15 cm

Explanation:

AGiven, $f_{0} + f_{e} = 30 cm$
And,
$\begin{array}{r} m = \frac{f_{0}}{f_{e}} = 2 \\ f_{0} = 2 f_{e} \end{array}$
Putting the value $f_{0}$ in equation (i), we get-
$\begin{array}{r} 2 f_{e} + f_{e} = 30 \\ f_{e} = \frac{30}{3} = 10 cm \end{array}$
Then,
$f_{0} = 2 f_{c} = 2 \times 10$
So, $f_{0} = 20 cm$

JEE Main-28.07.2022

Ray Optics

282354 The power of a lens (biconvex) is $1.25 m^{- 1}$ in particular medium. Refractive index of the lens is $1.5$ and radii of curvature are $20 cm$ and 40 cm respectively. The refractive index of surrounding medium:

1 1.0

2

\frac{9}{7}

3

\frac{3}{2}

4

\frac{4}{3}

Explanation:

B: Given, Power $(P) = 1.25 m^{- 1}$
Refractive index of lens $(μ_{g}) = 1.5$
Radius of curvature are, $R_{1} = 20 cm & R_{2} = - 40 cm$ Using lens maker formula-
$\begin{array}{r} P = \frac{1}{f} = (\frac{μ_{g}}{μ_{m}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{1}{0.2} + \frac{1}{0.4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{10}{2} + \frac{10}{4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (5 + \frac{5}{2}) \\ 1.25 = (\frac{1.5}{μ_{m}} - 1) (\frac{15}{2}) \\ \frac{2 \times 125}{15 \times 100} = (\frac{1.5}{μ_{m}} - 1) \\ \frac{2 \times 5}{15 \times 4} + 1 = \frac{3}{2 μ_{m}} \\ \frac{1}{3 \times 2} + 1 = \frac{3}{2 μ_{m}} \\ \frac{7}{6} = \frac{3}{2 μ} \\ μ_{m} = \frac{9}{7} \\ x \end{array}$

JEE Main-28.07.2022

Ray Optics

282351 When a drop of oil is spread on a water surface, it displays beautiful colours in day light because of:

1 dispersion of light

2 reflection of light

3 polarization of light

4 interference of light

Assam CEE-2017

Ray Optics

282352 The plane face of a Plano convex lens is silvered. If $μ$ be the refractive index and $R$ is the radius of curvature of curved surface, then system will behave like a concave mirror of curvature

1

μ R

2

R^{2} / μ

3

R / (μ - 1)

4

[(μ + 1) / (μ - 1)] R

Explanation:

C: Consider a parallel beam of light incident on convex surface,
$\begin{array}{r} \frac{μ}{v_{1}} - \frac{1}{- \infty} = \frac{μ - 1}{R} \\ v_{1} = \frac{μ R}{μ - 1} \end{array}$
This image gets reflected from the mirror to form an image at distance ${\frac{μ R}{μ - 1}}$ to the left of lens.
Refraction again from convex surface-
$\begin{array}{r} \frac{1}{V_{2}} - \frac{μ}{\frac{μ R}{μ - 1}} = \frac{1 - μ}{- R} \\ \frac{1}{V_{2}} = \frac{2 (μ - 1)}{R} \\ \frac{1}{f} = \frac{2 (μ - 1)}{R} \\ f = \frac{R}{2 (μ - 1)} \\ R^{'} = 2 f = \frac{R}{μ - 1} \end{array}$
(left of side)
Hence, $R^{'} = \frac{R}{μ - 1}$

JIPMER-2017

Ray Optics

282353 In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30 cm$ . The focal length of the objective, when the angular magnification of the telescope is 2 , will be:

1

20 cm

2

30 cm

3

10 cm

4

15 cm

Explanation:

AGiven, $f_{0} + f_{e} = 30 cm$
And,
$\begin{array}{r} m = \frac{f_{0}}{f_{e}} = 2 \\ f_{0} = 2 f_{e} \end{array}$
Putting the value $f_{0}$ in equation (i), we get-
$\begin{array}{r} 2 f_{e} + f_{e} = 30 \\ f_{e} = \frac{30}{3} = 10 cm \end{array}$
Then,
$f_{0} = 2 f_{c} = 2 \times 10$
So, $f_{0} = 20 cm$

JEE Main-28.07.2022

Ray Optics

282354 The power of a lens (biconvex) is $1.25 m^{- 1}$ in particular medium. Refractive index of the lens is $1.5$ and radii of curvature are $20 cm$ and 40 cm respectively. The refractive index of surrounding medium:

1 1.0

2

\frac{9}{7}

3

\frac{3}{2}

4

\frac{4}{3}

Explanation:

B: Given, Power $(P) = 1.25 m^{- 1}$
Refractive index of lens $(μ_{g}) = 1.5$
Radius of curvature are, $R_{1} = 20 cm & R_{2} = - 40 cm$ Using lens maker formula-
$\begin{array}{r} P = \frac{1}{f} = (\frac{μ_{g}}{μ_{m}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{1}{0.2} + \frac{1}{0.4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (\frac{10}{2} + \frac{10}{4}) \\ 1.25 = \frac{1}{f} = (\frac{1.5}{μ_{m}} - 1) (5 + \frac{5}{2}) \\ 1.25 = (\frac{1.5}{μ_{m}} - 1) (\frac{15}{2}) \\ \frac{2 \times 125}{15 \times 100} = (\frac{1.5}{μ_{m}} - 1) \\ \frac{2 \times 5}{15 \times 4} + 1 = \frac{3}{2 μ_{m}} \\ \frac{1}{3 \times 2} + 1 = \frac{3}{2 μ_{m}} \\ \frac{7}{6} = \frac{3}{2 μ} \\ μ_{m} = \frac{9}{7} \\ x \end{array}$

JEE Main-28.07.2022

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