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Ray Optics

282337 A convex lens of focal length $25 cm$ and made of glass with refractive index 1.5 is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water $= \frac{4}{3}$ ]

1

100 cm

2

37.5 cm

3

75 cm

4

12.5 cm

Explanation:

C Given that,
Focal length $(f_{1}) = 25 cm$
Refractive index of glass $(μ_{g}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know that,
$\frac{1}{f_{1}} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Again- $\frac{1}{f_{2}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
On dividing equation (i) by equation (ii)
$\begin{array}{r} \frac{f_{2}}{f_{1}} = \frac{(μ_{g} - 1)}{(\frac{μ_{g}}{μ_{w}} - 1)} \\ \frac{f_{2}}{25} = \frac{(1.5 - 1)}{(\frac{1.5}{4 / 3} - 1)} \\ f_{2} = \frac{0.5 \times 25}{0.125} = 100 cm \end{array}$
Hence, change in focal length $(Δ f) = f_{2} - f_{1}$
$= 100 - 25 = 75 cm$

TS EAMCET 20.07.2022

Ray Optics

282338 Two convex lenses have focal lengths of $50 cm$ and $25 cm$ , respectively. If these two lenses are placed in contact, then the net power of this combination will be equal to

1 +2 dioptre

2 +6 dioptre

3 -6 dioptre

4 +3 dioptre

Explanation:

B: Given, $f_{1} = 50 cm = 0.5 m$
$f_{2} = 25 cm = 0.25 m$
Power of lens $= \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\begin{array}{r} P = \frac{1}{0.5} + \frac{1}{0.25} \\ P = 2 + 4 \\ P = + 6 D \end{array}$

UPSC NDA-04.09.2022

Ray Optics

282339 What is the focal length of a convex lens of focal length $30 cm$ in contact with a concave lens of focal length $10 cm$ ?

1

- 15 cm

2

- 40 cm

3

- 20 cm

4

- 30 cm

Explanation:

A: Given, $f_{1} = 30 cm, f_{2} = - 10 cm$
$\begin{array}{r} f = \frac{f_{1} \cdot f_{2}}{f_{1} + f_{2}} \\ f = \frac{30 \times (- 10)}{30 + (- 10)} \\ f = \frac{- 300}{20} \\ f = - 15 cm \end{array}$

GUJCET 18.04.2022

Ray Optics

282340 A lens is made of glass having an index of refraction 1.5. One side of the lens is flat and the other side is convex with a radius $R$ . If an object is placed $60 cm$ , towards the convex side of the lens, the image is formed at $120 cm$ on the other side of the lens. The value of $R$ is

1

20 cm

2

\frac{40}{3} cm

3

33 cm

4

18 cm

Explanation:

A: Given, $μ_{2} = 1.5, u = - 60 cm, v = 120 cm$ For focal length, using lens formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{f} = \frac{1}{120} + \frac{1}{60} \\ f = 40 cm \end{array}$
Now, $R_{1} = R & R_{2} = \infty$
Using lens maker formula,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{40} = (1.5 - 1) (\frac{1}{R} - \frac{1}{\infty}) \\ R = 40 \times 0.5 \\ R = 20 cm \end{array}$

TS EAMCET 18.07.2022

Ray Optics

282337 A convex lens of focal length $25 cm$ and made of glass with refractive index 1.5 is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water $= \frac{4}{3}$ ]

1

100 cm

2

37.5 cm

3

75 cm

4

12.5 cm

Explanation:

C Given that,
Focal length $(f_{1}) = 25 cm$
Refractive index of glass $(μ_{g}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know that,
$\frac{1}{f_{1}} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Again- $\frac{1}{f_{2}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
On dividing equation (i) by equation (ii)
$\begin{array}{r} \frac{f_{2}}{f_{1}} = \frac{(μ_{g} - 1)}{(\frac{μ_{g}}{μ_{w}} - 1)} \\ \frac{f_{2}}{25} = \frac{(1.5 - 1)}{(\frac{1.5}{4 / 3} - 1)} \\ f_{2} = \frac{0.5 \times 25}{0.125} = 100 cm \end{array}$
Hence, change in focal length $(Δ f) = f_{2} - f_{1}$
$= 100 - 25 = 75 cm$

TS EAMCET 20.07.2022

Ray Optics

282338 Two convex lenses have focal lengths of $50 cm$ and $25 cm$ , respectively. If these two lenses are placed in contact, then the net power of this combination will be equal to

1 +2 dioptre

2 +6 dioptre

3 -6 dioptre

4 +3 dioptre

Explanation:

B: Given, $f_{1} = 50 cm = 0.5 m$
$f_{2} = 25 cm = 0.25 m$
Power of lens $= \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\begin{array}{r} P = \frac{1}{0.5} + \frac{1}{0.25} \\ P = 2 + 4 \\ P = + 6 D \end{array}$

UPSC NDA-04.09.2022

Ray Optics

282339 What is the focal length of a convex lens of focal length $30 cm$ in contact with a concave lens of focal length $10 cm$ ?

1

- 15 cm

2

- 40 cm

3

- 20 cm

4

- 30 cm

Explanation:

A: Given, $f_{1} = 30 cm, f_{2} = - 10 cm$
$\begin{array}{r} f = \frac{f_{1} \cdot f_{2}}{f_{1} + f_{2}} \\ f = \frac{30 \times (- 10)}{30 + (- 10)} \\ f = \frac{- 300}{20} \\ f = - 15 cm \end{array}$

GUJCET 18.04.2022

Ray Optics

282340 A lens is made of glass having an index of refraction 1.5. One side of the lens is flat and the other side is convex with a radius $R$ . If an object is placed $60 cm$ , towards the convex side of the lens, the image is formed at $120 cm$ on the other side of the lens. The value of $R$ is

1

20 cm

2

\frac{40}{3} cm

3

33 cm

4

18 cm

Explanation:

A: Given, $μ_{2} = 1.5, u = - 60 cm, v = 120 cm$ For focal length, using lens formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{f} = \frac{1}{120} + \frac{1}{60} \\ f = 40 cm \end{array}$
Now, $R_{1} = R & R_{2} = \infty$
Using lens maker formula,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{40} = (1.5 - 1) (\frac{1}{R} - \frac{1}{\infty}) \\ R = 40 \times 0.5 \\ R = 20 cm \end{array}$

TS EAMCET 18.07.2022

Ray Optics

282337 A convex lens of focal length $25 cm$ and made of glass with refractive index 1.5 is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water $= \frac{4}{3}$ ]

1

100 cm

2

37.5 cm

3

75 cm

4

12.5 cm

Explanation:

C Given that,
Focal length $(f_{1}) = 25 cm$
Refractive index of glass $(μ_{g}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know that,
$\frac{1}{f_{1}} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Again- $\frac{1}{f_{2}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
On dividing equation (i) by equation (ii)
$\begin{array}{r} \frac{f_{2}}{f_{1}} = \frac{(μ_{g} - 1)}{(\frac{μ_{g}}{μ_{w}} - 1)} \\ \frac{f_{2}}{25} = \frac{(1.5 - 1)}{(\frac{1.5}{4 / 3} - 1)} \\ f_{2} = \frac{0.5 \times 25}{0.125} = 100 cm \end{array}$
Hence, change in focal length $(Δ f) = f_{2} - f_{1}$
$= 100 - 25 = 75 cm$

TS EAMCET 20.07.2022

Ray Optics

282338 Two convex lenses have focal lengths of $50 cm$ and $25 cm$ , respectively. If these two lenses are placed in contact, then the net power of this combination will be equal to

1 +2 dioptre

2 +6 dioptre

3 -6 dioptre

4 +3 dioptre

Explanation:

B: Given, $f_{1} = 50 cm = 0.5 m$
$f_{2} = 25 cm = 0.25 m$
Power of lens $= \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\begin{array}{r} P = \frac{1}{0.5} + \frac{1}{0.25} \\ P = 2 + 4 \\ P = + 6 D \end{array}$

UPSC NDA-04.09.2022

Ray Optics

282339 What is the focal length of a convex lens of focal length $30 cm$ in contact with a concave lens of focal length $10 cm$ ?

1

- 15 cm

2

- 40 cm

3

- 20 cm

4

- 30 cm

Explanation:

A: Given, $f_{1} = 30 cm, f_{2} = - 10 cm$
$\begin{array}{r} f = \frac{f_{1} \cdot f_{2}}{f_{1} + f_{2}} \\ f = \frac{30 \times (- 10)}{30 + (- 10)} \\ f = \frac{- 300}{20} \\ f = - 15 cm \end{array}$

GUJCET 18.04.2022

Ray Optics

282340 A lens is made of glass having an index of refraction 1.5. One side of the lens is flat and the other side is convex with a radius $R$ . If an object is placed $60 cm$ , towards the convex side of the lens, the image is formed at $120 cm$ on the other side of the lens. The value of $R$ is

1

20 cm

2

\frac{40}{3} cm

3

33 cm

4

18 cm

Explanation:

A: Given, $μ_{2} = 1.5, u = - 60 cm, v = 120 cm$ For focal length, using lens formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{f} = \frac{1}{120} + \frac{1}{60} \\ f = 40 cm \end{array}$
Now, $R_{1} = R & R_{2} = \infty$
Using lens maker formula,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{40} = (1.5 - 1) (\frac{1}{R} - \frac{1}{\infty}) \\ R = 40 \times 0.5 \\ R = 20 cm \end{array}$

TS EAMCET 18.07.2022

Ray Optics

282337 A convex lens of focal length $25 cm$ and made of glass with refractive index 1.5 is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water $= \frac{4}{3}$ ]

1

100 cm

2

37.5 cm

3

75 cm

4

12.5 cm

Explanation:

C Given that,
Focal length $(f_{1}) = 25 cm$
Refractive index of glass $(μ_{g}) = 1.5$
Refractive index of water $(μ_{w}) = 4 / 3$
We know that,
$\frac{1}{f_{1}} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Again- $\frac{1}{f_{2}} = (\frac{μ_{g}}{μ_{w}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
On dividing equation (i) by equation (ii)
$\begin{array}{r} \frac{f_{2}}{f_{1}} = \frac{(μ_{g} - 1)}{(\frac{μ_{g}}{μ_{w}} - 1)} \\ \frac{f_{2}}{25} = \frac{(1.5 - 1)}{(\frac{1.5}{4 / 3} - 1)} \\ f_{2} = \frac{0.5 \times 25}{0.125} = 100 cm \end{array}$
Hence, change in focal length $(Δ f) = f_{2} - f_{1}$
$= 100 - 25 = 75 cm$

TS EAMCET 20.07.2022

Ray Optics

282338 Two convex lenses have focal lengths of $50 cm$ and $25 cm$ , respectively. If these two lenses are placed in contact, then the net power of this combination will be equal to

1 +2 dioptre

2 +6 dioptre

3 -6 dioptre

4 +3 dioptre

Explanation:

B: Given, $f_{1} = 50 cm = 0.5 m$
$f_{2} = 25 cm = 0.25 m$
Power of lens $= \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$\begin{array}{r} P = \frac{1}{0.5} + \frac{1}{0.25} \\ P = 2 + 4 \\ P = + 6 D \end{array}$

UPSC NDA-04.09.2022

Ray Optics

282339 What is the focal length of a convex lens of focal length $30 cm$ in contact with a concave lens of focal length $10 cm$ ?

1

- 15 cm

2

- 40 cm

3

- 20 cm

4

- 30 cm

Explanation:

A: Given, $f_{1} = 30 cm, f_{2} = - 10 cm$
$\begin{array}{r} f = \frac{f_{1} \cdot f_{2}}{f_{1} + f_{2}} \\ f = \frac{30 \times (- 10)}{30 + (- 10)} \\ f = \frac{- 300}{20} \\ f = - 15 cm \end{array}$

GUJCET 18.04.2022

Ray Optics

282340 A lens is made of glass having an index of refraction 1.5. One side of the lens is flat and the other side is convex with a radius $R$ . If an object is placed $60 cm$ , towards the convex side of the lens, the image is formed at $120 cm$ on the other side of the lens. The value of $R$ is

1

20 cm

2

\frac{40}{3} cm

3

33 cm

4

18 cm

Explanation:

A: Given, $μ_{2} = 1.5, u = - 60 cm, v = 120 cm$ For focal length, using lens formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{f} = \frac{1}{120} + \frac{1}{60} \\ f = 40 cm \end{array}$
Now, $R_{1} = R & R_{2} = \infty$
Using lens maker formula,
$\begin{array}{r} \frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{40} = (1.5 - 1) (\frac{1}{R} - \frac{1}{\infty}) \\ R = 40 \times 0.5 \\ R = 20 cm \end{array}$

TS EAMCET 18.07.2022