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Ray Optics

282460 A converging lens forms a real image $I$ on its optic axis. A rectangular glass slab of refractive inde $x μ$ and thickness $t$ is introduced between the lens and the image $I$ . The image I will move

1 away from the lens by

t [1 - \frac{1}{μ}]

2 towards the lens by

t [1 - \frac{1}{μ}]

3 away from the lens by

t (μ - 1)

4 towards the lens by

t (μ - 1)

Explanation:

A: Due to insertion of slab, the optical path increased by $\frac{t}{μ}$ , where $t$ is thickness of slab. Therefore converging point will shift away by -
$(t - \frac{t}{μ}) = t (1 - \frac{1}{μ})$

CG PET- 2017

Ray Optics

282461 A point object is placed at a distance of $20 cm$ from a thin Plano-convex lens of focal length 15 $cm$ , if the plane surface is silvered. The image will form at

1

60 cm

left of

AB

2

30 cm

left of

AB

3

12 cm

left of

A B

4

60 cm

right of

A B

Explanation:

C: Given that, $u = - 20 cm$
Focus length of plano convex lens is $\frac{f}{2} = \frac{- 15}{2}$
Using mirror formula -
$\begin{array}{r} \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} + \frac{1}{- 20} = \frac{2}{- 15} \\ \frac{1}{v} = \frac{- 2}{15} + \frac{1}{20} \\ \frac{1}{v} = \frac{- 8 + 3}{60} \\ v = - 12 cm \end{array}$
$12 cm$ left of $AB$ .

Manipal UGET-2014

Ray Optics

282462 The Plano-convex lens of focal length $20 cm$ and $30 cm$ are placed together to form a double convex lens, the final focal length will be

1

12 cm

2

60 cm

3

20 cm

4

30 cm

Explanation:

A: Given that,
Focal length, $f_{1} = 20 cm$
$f_{2} = 30 cm$
We know that equivalent focal length,
$\begin{aligned} \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ \frac{1}{f} = \frac{1}{20} + \frac{1}{30} \\ \frac{1}{f} = \frac{5}{60} \\ F = 12 cm \end{aligned}$

Manipal UGET-2013

Ray Optics

282463 Power of a lens is - 4D and for second lens, power is $+ 2 D$ ; the total power of the couple is

1

- 2 D

2

6 D

3

- 6 D

4

- 8 D

Explanation:

A: Given that,
Power $(P_{1}) = - 4 D$ , Power $(P_{2}) = + 2 D$
We know that,
Combination of power of lens
$\begin{array}{r} P = P_{1} + P_{2} \\ P = - 4 + 2 \\ P = - 2 D \end{array}$

Manipal UGET-2012

Ray Optics

282460 A converging lens forms a real image $I$ on its optic axis. A rectangular glass slab of refractive inde $x μ$ and thickness $t$ is introduced between the lens and the image $I$ . The image I will move

1 away from the lens by

t [1 - \frac{1}{μ}]

2 towards the lens by

t [1 - \frac{1}{μ}]

3 away from the lens by

t (μ - 1)

4 towards the lens by

t (μ - 1)

Explanation:

A: Due to insertion of slab, the optical path increased by $\frac{t}{μ}$ , where $t$ is thickness of slab. Therefore converging point will shift away by -
$(t - \frac{t}{μ}) = t (1 - \frac{1}{μ})$

CG PET- 2017

Ray Optics

282461 A point object is placed at a distance of $20 cm$ from a thin Plano-convex lens of focal length 15 $cm$ , if the plane surface is silvered. The image will form at

1

60 cm

left of

AB

2

30 cm

left of

AB

3

12 cm

left of

A B

4

60 cm

right of

A B

Explanation:

C: Given that, $u = - 20 cm$
Focus length of plano convex lens is $\frac{f}{2} = \frac{- 15}{2}$
Using mirror formula -
$\begin{array}{r} \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} + \frac{1}{- 20} = \frac{2}{- 15} \\ \frac{1}{v} = \frac{- 2}{15} + \frac{1}{20} \\ \frac{1}{v} = \frac{- 8 + 3}{60} \\ v = - 12 cm \end{array}$
$12 cm$ left of $AB$ .

Manipal UGET-2014

Ray Optics

282462 The Plano-convex lens of focal length $20 cm$ and $30 cm$ are placed together to form a double convex lens, the final focal length will be

1

12 cm

2

60 cm

3

20 cm

4

30 cm

Explanation:

A: Given that,
Focal length, $f_{1} = 20 cm$
$f_{2} = 30 cm$
We know that equivalent focal length,
$\begin{aligned} \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ \frac{1}{f} = \frac{1}{20} + \frac{1}{30} \\ \frac{1}{f} = \frac{5}{60} \\ F = 12 cm \end{aligned}$

Manipal UGET-2013

Ray Optics

282463 Power of a lens is - 4D and for second lens, power is $+ 2 D$ ; the total power of the couple is

1

- 2 D

2

6 D

3

- 6 D

4

- 8 D

Explanation:

A: Given that,
Power $(P_{1}) = - 4 D$ , Power $(P_{2}) = + 2 D$
We know that,
Combination of power of lens
$\begin{array}{r} P = P_{1} + P_{2} \\ P = - 4 + 2 \\ P = - 2 D \end{array}$

Manipal UGET-2012

Ray Optics

282460 A converging lens forms a real image $I$ on its optic axis. A rectangular glass slab of refractive inde $x μ$ and thickness $t$ is introduced between the lens and the image $I$ . The image I will move

1 away from the lens by

t [1 - \frac{1}{μ}]

2 towards the lens by

t [1 - \frac{1}{μ}]

3 away from the lens by

t (μ - 1)

4 towards the lens by

t (μ - 1)

Explanation:

A: Due to insertion of slab, the optical path increased by $\frac{t}{μ}$ , where $t$ is thickness of slab. Therefore converging point will shift away by -
$(t - \frac{t}{μ}) = t (1 - \frac{1}{μ})$

CG PET- 2017

Ray Optics

282461 A point object is placed at a distance of $20 cm$ from a thin Plano-convex lens of focal length 15 $cm$ , if the plane surface is silvered. The image will form at

1

60 cm

left of

AB

2

30 cm

left of

AB

3

12 cm

left of

A B

4

60 cm

right of

A B

Explanation:

C: Given that, $u = - 20 cm$
Focus length of plano convex lens is $\frac{f}{2} = \frac{- 15}{2}$
Using mirror formula -
$\begin{array}{r} \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} + \frac{1}{- 20} = \frac{2}{- 15} \\ \frac{1}{v} = \frac{- 2}{15} + \frac{1}{20} \\ \frac{1}{v} = \frac{- 8 + 3}{60} \\ v = - 12 cm \end{array}$
$12 cm$ left of $AB$ .

Manipal UGET-2014

Ray Optics

282462 The Plano-convex lens of focal length $20 cm$ and $30 cm$ are placed together to form a double convex lens, the final focal length will be

1

12 cm

2

60 cm

3

20 cm

4

30 cm

Explanation:

A: Given that,
Focal length, $f_{1} = 20 cm$
$f_{2} = 30 cm$
We know that equivalent focal length,
$\begin{aligned} \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ \frac{1}{f} = \frac{1}{20} + \frac{1}{30} \\ \frac{1}{f} = \frac{5}{60} \\ F = 12 cm \end{aligned}$

Manipal UGET-2013

Ray Optics

282463 Power of a lens is - 4D and for second lens, power is $+ 2 D$ ; the total power of the couple is

1

- 2 D

2

6 D

3

- 6 D

4

- 8 D

Explanation:

A: Given that,
Power $(P_{1}) = - 4 D$ , Power $(P_{2}) = + 2 D$
We know that,
Combination of power of lens
$\begin{array}{r} P = P_{1} + P_{2} \\ P = - 4 + 2 \\ P = - 2 D \end{array}$

Manipal UGET-2012

Ray Optics

282460 A converging lens forms a real image $I$ on its optic axis. A rectangular glass slab of refractive inde $x μ$ and thickness $t$ is introduced between the lens and the image $I$ . The image I will move

1 away from the lens by

t [1 - \frac{1}{μ}]

2 towards the lens by

t [1 - \frac{1}{μ}]

3 away from the lens by

t (μ - 1)

4 towards the lens by

t (μ - 1)

Explanation:

A: Due to insertion of slab, the optical path increased by $\frac{t}{μ}$ , where $t$ is thickness of slab. Therefore converging point will shift away by -
$(t - \frac{t}{μ}) = t (1 - \frac{1}{μ})$

CG PET- 2017

Ray Optics

282461 A point object is placed at a distance of $20 cm$ from a thin Plano-convex lens of focal length 15 $cm$ , if the plane surface is silvered. The image will form at

1

60 cm

left of

AB

2

30 cm

left of

AB

3

12 cm

left of

A B

4

60 cm

right of

A B

Explanation:

C: Given that, $u = - 20 cm$
Focus length of plano convex lens is $\frac{f}{2} = \frac{- 15}{2}$
Using mirror formula -
$\begin{array}{r} \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{v} + \frac{1}{- 20} = \frac{2}{- 15} \\ \frac{1}{v} = \frac{- 2}{15} + \frac{1}{20} \\ \frac{1}{v} = \frac{- 8 + 3}{60} \\ v = - 12 cm \end{array}$
$12 cm$ left of $AB$ .

Manipal UGET-2014

Ray Optics

282462 The Plano-convex lens of focal length $20 cm$ and $30 cm$ are placed together to form a double convex lens, the final focal length will be

1

12 cm

2

60 cm

3

20 cm

4

30 cm

Explanation:

A: Given that,
Focal length, $f_{1} = 20 cm$
$f_{2} = 30 cm$
We know that equivalent focal length,
$\begin{aligned} \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ \frac{1}{f} = \frac{1}{20} + \frac{1}{30} \\ \frac{1}{f} = \frac{5}{60} \\ F = 12 cm \end{aligned}$

Manipal UGET-2013

Ray Optics

282463 Power of a lens is - 4D and for second lens, power is $+ 2 D$ ; the total power of the couple is

1

- 2 D

2

6 D

3

- 6 D

4

- 8 D

Explanation:

A: Given that,
Power $(P_{1}) = - 4 D$ , Power $(P_{2}) = + 2 D$
We know that,
Combination of power of lens
$\begin{array}{r} P = P_{1} + P_{2} \\ P = - 4 + 2 \\ P = - 2 D \end{array}$

Manipal UGET-2012

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