QBManager

Ray Optics

282380 A biconvex lens $(R_{1} = R_{2} = 20 cm)$ has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens $= 1.5$ )

1

- 20 cm

2

40 cm

3

20 cm

4

- 40 cm

Explanation:

D: Given, $R_{1} = R_{2} = 20 cm$
Focal length of lens $=$ focal length of concave mirror.
$R$ of concave mirror $=$ ?
$μ_{g} = 1.5$
We know that,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) \\ \frac{1}{f} = \frac{0.5 \times 2}{20} \\ f = 20 cm = focal length of mirror \end{array}$
Hence, Radius of curvature of concave mirror
$= 2 \times f = 2 \times (- 20) = - 40 cm$
(Negative sign for concave mirror)

МНT-CET 2020

Ray Optics

282363 The focal length of a biconvex lens, made of glass, of equal radii is $f$ . If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as $3 / 2$ and $4 / 3$ , respectively)

1

2 f

2

4 f

3

(5 / 3) f

4

(7 / 4) f

Explanation:

B: Radius are equal,
$\begin{array}{r} R_{1} = R_{2} = r \\ _{a} μ_{g} = \frac{3}{2},_{a} μ_{w} = \frac{4}{3} \end{array}$
According to lens makers formula,
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Here, $R_{1} = R$ and $R_{2} = - R$
$\begin{aligned} ∴ \frac{1}{f} & = (\frac{3}{2} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{2} \times \frac{2}{R} = \frac{1}{R} \\ = R \end{aligned}$
When the lens is dipped in the water,
$\begin{array}{r} \frac{1}{f_{w}} = (\frac{_{a} μ_{g}}{_{a}} - 1) (\frac{1}{w} + \frac{1}{R}) \\ \frac{1}{f_{w}} = (\frac{3 / 2 - 1}{4 / 3 - 1}) \times \frac{2}{R} = (\frac{9}{8} - 1) \times \frac{2}{R} \\ \frac{1}{f_{w}} = \frac{1}{8} \times \frac{2}{R} = \frac{1}{4 R} \\ f_{w} = 4 f \end{array}$

TS EAMCET 06.08.2021

Ray Optics

282364 Two lenses of power $- 1.6 D$ and +2.1D respectively are placed in contact. The focal length of the combinations is

1

100 cm

2

200 cm

3

160 cm

4

210 cm

Explanation:

B: Given, $P_{1} = - 1.6 D, P_{2} = + 2.1 D$
Combined Power, $P = P_{1} + P_{2}$
$\begin{array}{r} P = - 1.6 + 2.1 \\ P = + 0.5 D \end{array}$
Focal length, $f = \frac{1}{P} = \frac{1}{0.5} = 2 m$ or
$f = 200 cm$

TS EAMCET 04.08.2021

Ray Optics

282365 Find equivalent focal length due to combination of two convex lens are in contact having a focal length both of them $30 cm$ .

1

15 cm

2

30 cm

3

20 cm

4

40 cm

Explanation:

A: $f_{1} = f_{2} = 30 cm$
Combination of lenses -
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{F} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{F} = \frac{2}{30}$
$F = 15 cm$

GUJCET-PCE- 2021

Ray Optics

282366 The Radii of curvature of the faces of a double convex lens are $10 cm$ and $15 cm$ . Its focal length is $12 cm$ . What is the refractive index of material of lens?

1 1.33

2 1.62

3 1.50

4 2.42

Explanation:

C: Given,
Radius of curvature of first face $(R_{1}) = 10 cm$
Radius of curvature of second face $(R_{2}) = - 15 cm$
Focal length $(f) = 12 cm$
Using formula -
$\begin{array}{r} \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) \\ \frac{1}{12} = (μ - 1) (\frac{3 + 2}{30}) \\ μ - 1 = \frac{6}{12} = \frac{1}{2} \\ μ = \frac{1}{2} + 1 = 3 / 2 \\ μ = 1.5 \end{array}$
or
$μ = 1.5$

GUJCET-PCE- 2021

Ray Optics

282380 A biconvex lens $(R_{1} = R_{2} = 20 cm)$ has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens $= 1.5$ )

1

- 20 cm

2

40 cm

3

20 cm

4

- 40 cm

Explanation:

D: Given, $R_{1} = R_{2} = 20 cm$
Focal length of lens $=$ focal length of concave mirror.
$R$ of concave mirror $=$ ?
$μ_{g} = 1.5$
We know that,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) \\ \frac{1}{f} = \frac{0.5 \times 2}{20} \\ f = 20 cm = focal length of mirror \end{array}$
Hence, Radius of curvature of concave mirror
$= 2 \times f = 2 \times (- 20) = - 40 cm$
(Negative sign for concave mirror)

МНT-CET 2020

Ray Optics

282363 The focal length of a biconvex lens, made of glass, of equal radii is $f$ . If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as $3 / 2$ and $4 / 3$ , respectively)

1

2 f

2

4 f

3

(5 / 3) f

4

(7 / 4) f

Explanation:

B: Radius are equal,
$\begin{array}{r} R_{1} = R_{2} = r \\ _{a} μ_{g} = \frac{3}{2},_{a} μ_{w} = \frac{4}{3} \end{array}$
According to lens makers formula,
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Here, $R_{1} = R$ and $R_{2} = - R$
$\begin{aligned} ∴ \frac{1}{f} & = (\frac{3}{2} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{2} \times \frac{2}{R} = \frac{1}{R} \\ = R \end{aligned}$
When the lens is dipped in the water,
$\begin{array}{r} \frac{1}{f_{w}} = (\frac{_{a} μ_{g}}{_{a}} - 1) (\frac{1}{w} + \frac{1}{R}) \\ \frac{1}{f_{w}} = (\frac{3 / 2 - 1}{4 / 3 - 1}) \times \frac{2}{R} = (\frac{9}{8} - 1) \times \frac{2}{R} \\ \frac{1}{f_{w}} = \frac{1}{8} \times \frac{2}{R} = \frac{1}{4 R} \\ f_{w} = 4 f \end{array}$

TS EAMCET 06.08.2021

Ray Optics

282364 Two lenses of power $- 1.6 D$ and +2.1D respectively are placed in contact. The focal length of the combinations is

1

100 cm

2

200 cm

3

160 cm

4

210 cm

Explanation:

B: Given, $P_{1} = - 1.6 D, P_{2} = + 2.1 D$
Combined Power, $P = P_{1} + P_{2}$
$\begin{array}{r} P = - 1.6 + 2.1 \\ P = + 0.5 D \end{array}$
Focal length, $f = \frac{1}{P} = \frac{1}{0.5} = 2 m$ or
$f = 200 cm$

TS EAMCET 04.08.2021

Ray Optics

282365 Find equivalent focal length due to combination of two convex lens are in contact having a focal length both of them $30 cm$ .

1

15 cm

2

30 cm

3

20 cm

4

40 cm

Explanation:

A: $f_{1} = f_{2} = 30 cm$
Combination of lenses -
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{F} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{F} = \frac{2}{30}$
$F = 15 cm$

GUJCET-PCE- 2021

Ray Optics

282366 The Radii of curvature of the faces of a double convex lens are $10 cm$ and $15 cm$ . Its focal length is $12 cm$ . What is the refractive index of material of lens?

1 1.33

2 1.62

3 1.50

4 2.42

Explanation:

C: Given,
Radius of curvature of first face $(R_{1}) = 10 cm$
Radius of curvature of second face $(R_{2}) = - 15 cm$
Focal length $(f) = 12 cm$
Using formula -
$\begin{array}{r} \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) \\ \frac{1}{12} = (μ - 1) (\frac{3 + 2}{30}) \\ μ - 1 = \frac{6}{12} = \frac{1}{2} \\ μ = \frac{1}{2} + 1 = 3 / 2 \\ μ = 1.5 \end{array}$
or
$μ = 1.5$

GUJCET-PCE- 2021

Ray Optics

282380 A biconvex lens $(R_{1} = R_{2} = 20 cm)$ has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens $= 1.5$ )

1

- 20 cm

2

40 cm

3

20 cm

4

- 40 cm

Explanation:

D: Given, $R_{1} = R_{2} = 20 cm$
Focal length of lens $=$ focal length of concave mirror.
$R$ of concave mirror $=$ ?
$μ_{g} = 1.5$
We know that,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) \\ \frac{1}{f} = \frac{0.5 \times 2}{20} \\ f = 20 cm = focal length of mirror \end{array}$
Hence, Radius of curvature of concave mirror
$= 2 \times f = 2 \times (- 20) = - 40 cm$
(Negative sign for concave mirror)

МНT-CET 2020

Ray Optics

282363 The focal length of a biconvex lens, made of glass, of equal radii is $f$ . If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as $3 / 2$ and $4 / 3$ , respectively)

1

2 f

2

4 f

3

(5 / 3) f

4

(7 / 4) f

Explanation:

B: Radius are equal,
$\begin{array}{r} R_{1} = R_{2} = r \\ _{a} μ_{g} = \frac{3}{2},_{a} μ_{w} = \frac{4}{3} \end{array}$
According to lens makers formula,
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Here, $R_{1} = R$ and $R_{2} = - R$
$\begin{aligned} ∴ \frac{1}{f} & = (\frac{3}{2} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{2} \times \frac{2}{R} = \frac{1}{R} \\ = R \end{aligned}$
When the lens is dipped in the water,
$\begin{array}{r} \frac{1}{f_{w}} = (\frac{_{a} μ_{g}}{_{a}} - 1) (\frac{1}{w} + \frac{1}{R}) \\ \frac{1}{f_{w}} = (\frac{3 / 2 - 1}{4 / 3 - 1}) \times \frac{2}{R} = (\frac{9}{8} - 1) \times \frac{2}{R} \\ \frac{1}{f_{w}} = \frac{1}{8} \times \frac{2}{R} = \frac{1}{4 R} \\ f_{w} = 4 f \end{array}$

TS EAMCET 06.08.2021

Ray Optics

282364 Two lenses of power $- 1.6 D$ and +2.1D respectively are placed in contact. The focal length of the combinations is

1

100 cm

2

200 cm

3

160 cm

4

210 cm

Explanation:

B: Given, $P_{1} = - 1.6 D, P_{2} = + 2.1 D$
Combined Power, $P = P_{1} + P_{2}$
$\begin{array}{r} P = - 1.6 + 2.1 \\ P = + 0.5 D \end{array}$
Focal length, $f = \frac{1}{P} = \frac{1}{0.5} = 2 m$ or
$f = 200 cm$

TS EAMCET 04.08.2021

Ray Optics

282365 Find equivalent focal length due to combination of two convex lens are in contact having a focal length both of them $30 cm$ .

1

15 cm

2

30 cm

3

20 cm

4

40 cm

Explanation:

A: $f_{1} = f_{2} = 30 cm$
Combination of lenses -
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{F} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{F} = \frac{2}{30}$
$F = 15 cm$

GUJCET-PCE- 2021

Ray Optics

282366 The Radii of curvature of the faces of a double convex lens are $10 cm$ and $15 cm$ . Its focal length is $12 cm$ . What is the refractive index of material of lens?

1 1.33

2 1.62

3 1.50

4 2.42

Explanation:

C: Given,
Radius of curvature of first face $(R_{1}) = 10 cm$
Radius of curvature of second face $(R_{2}) = - 15 cm$
Focal length $(f) = 12 cm$
Using formula -
$\begin{array}{r} \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) \\ \frac{1}{12} = (μ - 1) (\frac{3 + 2}{30}) \\ μ - 1 = \frac{6}{12} = \frac{1}{2} \\ μ = \frac{1}{2} + 1 = 3 / 2 \\ μ = 1.5 \end{array}$
or
$μ = 1.5$

GUJCET-PCE- 2021

Ray Optics

282380 A biconvex lens $(R_{1} = R_{2} = 20 cm)$ has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens $= 1.5$ )

1

- 20 cm

2

40 cm

3

20 cm

4

- 40 cm

Explanation:

D: Given, $R_{1} = R_{2} = 20 cm$
Focal length of lens $=$ focal length of concave mirror.
$R$ of concave mirror $=$ ?
$μ_{g} = 1.5$
We know that,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) \\ \frac{1}{f} = \frac{0.5 \times 2}{20} \\ f = 20 cm = focal length of mirror \end{array}$
Hence, Radius of curvature of concave mirror
$= 2 \times f = 2 \times (- 20) = - 40 cm$
(Negative sign for concave mirror)

МНT-CET 2020

Ray Optics

282363 The focal length of a biconvex lens, made of glass, of equal radii is $f$ . If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as $3 / 2$ and $4 / 3$ , respectively)

1

2 f

2

4 f

3

(5 / 3) f

4

(7 / 4) f

Explanation:

B: Radius are equal,
$\begin{array}{r} R_{1} = R_{2} = r \\ _{a} μ_{g} = \frac{3}{2},_{a} μ_{w} = \frac{4}{3} \end{array}$
According to lens makers formula,
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Here, $R_{1} = R$ and $R_{2} = - R$
$\begin{aligned} ∴ \frac{1}{f} & = (\frac{3}{2} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{2} \times \frac{2}{R} = \frac{1}{R} \\ = R \end{aligned}$
When the lens is dipped in the water,
$\begin{array}{r} \frac{1}{f_{w}} = (\frac{_{a} μ_{g}}{_{a}} - 1) (\frac{1}{w} + \frac{1}{R}) \\ \frac{1}{f_{w}} = (\frac{3 / 2 - 1}{4 / 3 - 1}) \times \frac{2}{R} = (\frac{9}{8} - 1) \times \frac{2}{R} \\ \frac{1}{f_{w}} = \frac{1}{8} \times \frac{2}{R} = \frac{1}{4 R} \\ f_{w} = 4 f \end{array}$

TS EAMCET 06.08.2021

Ray Optics

282364 Two lenses of power $- 1.6 D$ and +2.1D respectively are placed in contact. The focal length of the combinations is

1

100 cm

2

200 cm

3

160 cm

4

210 cm

Explanation:

B: Given, $P_{1} = - 1.6 D, P_{2} = + 2.1 D$
Combined Power, $P = P_{1} + P_{2}$
$\begin{array}{r} P = - 1.6 + 2.1 \\ P = + 0.5 D \end{array}$
Focal length, $f = \frac{1}{P} = \frac{1}{0.5} = 2 m$ or
$f = 200 cm$

TS EAMCET 04.08.2021

Ray Optics

282365 Find equivalent focal length due to combination of two convex lens are in contact having a focal length both of them $30 cm$ .

1

15 cm

2

30 cm

3

20 cm

4

40 cm

Explanation:

A: $f_{1} = f_{2} = 30 cm$
Combination of lenses -
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{F} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{F} = \frac{2}{30}$
$F = 15 cm$

GUJCET-PCE- 2021

Ray Optics

282366 The Radii of curvature of the faces of a double convex lens are $10 cm$ and $15 cm$ . Its focal length is $12 cm$ . What is the refractive index of material of lens?

1 1.33

2 1.62

3 1.50

4 2.42

Explanation:

C: Given,
Radius of curvature of first face $(R_{1}) = 10 cm$
Radius of curvature of second face $(R_{2}) = - 15 cm$
Focal length $(f) = 12 cm$
Using formula -
$\begin{array}{r} \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) \\ \frac{1}{12} = (μ - 1) (\frac{3 + 2}{30}) \\ μ - 1 = \frac{6}{12} = \frac{1}{2} \\ μ = \frac{1}{2} + 1 = 3 / 2 \\ μ = 1.5 \end{array}$
or
$μ = 1.5$

GUJCET-PCE- 2021

Ray Optics

282380 A biconvex lens $(R_{1} = R_{2} = 20 cm)$ has focal length equal to focal length of concave mirror. The radius of curvature of concave mirror is (R.I. of glass lens $= 1.5$ )

1

- 20 cm

2

40 cm

3

20 cm

4

- 40 cm

Explanation:

D: Given, $R_{1} = R_{2} = 20 cm$
Focal length of lens $=$ focal length of concave mirror.
$R$ of concave mirror $=$ ?
$μ_{g} = 1.5$
We know that,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{f} = (1.5 - 1) (\frac{1}{20} + \frac{1}{20}) \\ \frac{1}{f} = \frac{0.5 \times 2}{20} \\ f = 20 cm = focal length of mirror \end{array}$
Hence, Radius of curvature of concave mirror
$= 2 \times f = 2 \times (- 20) = - 40 cm$
(Negative sign for concave mirror)

МНT-CET 2020

Ray Optics

282363 The focal length of a biconvex lens, made of glass, of equal radii is $f$ . If the lens is dipped in the water then the focal length becomes. (Take refractive index of glass and water as $3 / 2$ and $4 / 3$ , respectively)

1

2 f

2

4 f

3

(5 / 3) f

4

(7 / 4) f

Explanation:

B: Radius are equal,
$\begin{array}{r} R_{1} = R_{2} = r \\ _{a} μ_{g} = \frac{3}{2},_{a} μ_{w} = \frac{4}{3} \end{array}$
According to lens makers formula,
$∴ \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Here, $R_{1} = R$ and $R_{2} = - R$
$\begin{aligned} ∴ \frac{1}{f} & = (\frac{3}{2} - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1}{2} \times \frac{2}{R} = \frac{1}{R} \\ = R \end{aligned}$
When the lens is dipped in the water,
$\begin{array}{r} \frac{1}{f_{w}} = (\frac{_{a} μ_{g}}{_{a}} - 1) (\frac{1}{w} + \frac{1}{R}) \\ \frac{1}{f_{w}} = (\frac{3 / 2 - 1}{4 / 3 - 1}) \times \frac{2}{R} = (\frac{9}{8} - 1) \times \frac{2}{R} \\ \frac{1}{f_{w}} = \frac{1}{8} \times \frac{2}{R} = \frac{1}{4 R} \\ f_{w} = 4 f \end{array}$

TS EAMCET 06.08.2021

Ray Optics

282364 Two lenses of power $- 1.6 D$ and +2.1D respectively are placed in contact. The focal length of the combinations is

1

100 cm

2

200 cm

3

160 cm

4

210 cm

Explanation:

B: Given, $P_{1} = - 1.6 D, P_{2} = + 2.1 D$
Combined Power, $P = P_{1} + P_{2}$
$\begin{array}{r} P = - 1.6 + 2.1 \\ P = + 0.5 D \end{array}$
Focal length, $f = \frac{1}{P} = \frac{1}{0.5} = 2 m$ or
$f = 200 cm$

TS EAMCET 04.08.2021

Ray Optics

282365 Find equivalent focal length due to combination of two convex lens are in contact having a focal length both of them $30 cm$ .

1

15 cm

2

30 cm

3

20 cm

4

40 cm

Explanation:

A: $f_{1} = f_{2} = 30 cm$
Combination of lenses -
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\frac{1}{F} = \frac{1}{30} + \frac{1}{30}$
$\frac{1}{F} = \frac{2}{30}$
$F = 15 cm$

GUJCET-PCE- 2021

Ray Optics

282366 The Radii of curvature of the faces of a double convex lens are $10 cm$ and $15 cm$ . Its focal length is $12 cm$ . What is the refractive index of material of lens?

1 1.33

2 1.62

3 1.50

4 2.42

Explanation:

C: Given,
Radius of curvature of first face $(R_{1}) = 10 cm$
Radius of curvature of second face $(R_{2}) = - 15 cm$
Focal length $(f) = 12 cm$
Using formula -
$\begin{array}{r} \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) \\ \frac{1}{12} = (μ - 1) (\frac{3 + 2}{30}) \\ μ - 1 = \frac{6}{12} = \frac{1}{2} \\ μ = \frac{1}{2} + 1 = 3 / 2 \\ μ = 1.5 \end{array}$
or
$μ = 1.5$

GUJCET-PCE- 2021

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