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Ray Optics

282031 When an object is more along the principle axis of a concave mirror placed in air, the image coincides with the object if the object is $50 cm$ from the mirror. If the mirror is place at the depth of $20 cm$ in transparent medium, the image coincided with the object when the object is $40 cm$ from the mirror. The refractive index of the liquid is

1

\frac{5}{4}

2

\frac{4}{3}

3

\frac{3}{2}

4

\frac{5}{3}

Explanation:

C: Given,
When liquid is not present $u = - 50 cm$ and $v = - 50 cm$
Then, for focal length -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{50} - \frac{1}{50} \\ \frac{1}{f} = - \frac{2}{50} \\ f = - 25 cm \end{array}$
Lateral shift in liquid $= 20 cm$
Ray travel in liquid $= 40 cm$
We know,
Lateral shift $=$ Ray travel in liquid $(μ - 1)$
$\begin{array}{r} 20 = 40 (μ - 1) \\ \frac{1}{2} = μ - 1 \\ μ = 1 + \frac{1}{2} \end{array}$
Hence, the refractive index of the liquid-
$μ = \frac{3}{2}$

AP EAMCET (23.04.2019) Shift-I

Ray Optics

282032 For concave mirror, if the object is just beyond the focal length, the image formed is

1 real, inverted and magnified

2 real, inverted and diminshed

3 real, erect and magnified

4 real, erect and diminished

МНТ-СЕТ 2019

Ray Optics

282033 An object is placed at a distance of $40 cm$ from a concave mirror of focal length $15 cm$ . If the object is displaced through a distance of $20 cm$ towards the mirror, the displacement of the image will be

1

30 cm

towards the mirror

2

36 cm

away from the mirror

3

30 cm

away from the mirror

4

36 cm

toward the mirror

Explanation:

B: Case - I
Given, object distance $(u) = - 40 cm$
Focal length $(f) = - 15 cm$
From mirror formula
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{- 1}{15} = \frac{- 1}{40} + \frac{1}{v} \\ \frac{1}{V} = \frac{1}{40} - \frac{1}{15} \\ \frac{1}{v} = \frac{3 - 8}{120} \\ \frac{1}{V} = \frac{- 5}{120} \\ v = - 24 cm \end{array}$
Hence, image is formed in front of mirror at a distance $24 cm$ .
Case - II
Focus remains same
$f = - 15 cm$
Object new location from the mirror
$\begin{array}{r} u^{'} = u - 20 \\ u^{'} = 40 - 20 \\ u^{'} = 20 cm \end{array}$
Hence, object new position is $- 20 cm$ from the mirror formula
$\begin{aligned} \frac{1}{f} & = \frac{1}{u^{'}} + \frac{1}{v^{'}} \\ \frac{1}{v^{'}} & = \frac{1}{20} - \frac{1}{15} \\ \frac{1}{v^{'}} & = \frac{3 - 4}{60} \\ v^{'} & = - 60 cm \end{aligned}$
Hence, image move by
$\begin{array}{r} = v^{'} - v \\ = - 60 - (- 24) \\ = - 60 + 24 \\ = - 36 cm \end{array}$
Hence, image move by $36 cm$ away from the mirror.

NEET- 2018

Ray Optics

282034 Parallel rays of light entering a convex lens always converge at

1 Centre of curvature

2 The principle focus

3 Optical centre

4 Focal plane

SRM JEE-2018

Ray Optics

282031 When an object is more along the principle axis of a concave mirror placed in air, the image coincides with the object if the object is $50 cm$ from the mirror. If the mirror is place at the depth of $20 cm$ in transparent medium, the image coincided with the object when the object is $40 cm$ from the mirror. The refractive index of the liquid is

1

\frac{5}{4}

2

\frac{4}{3}

3

\frac{3}{2}

4

\frac{5}{3}

Explanation:

C: Given,
When liquid is not present $u = - 50 cm$ and $v = - 50 cm$
Then, for focal length -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{50} - \frac{1}{50} \\ \frac{1}{f} = - \frac{2}{50} \\ f = - 25 cm \end{array}$
Lateral shift in liquid $= 20 cm$
Ray travel in liquid $= 40 cm$
We know,
Lateral shift $=$ Ray travel in liquid $(μ - 1)$
$\begin{array}{r} 20 = 40 (μ - 1) \\ \frac{1}{2} = μ - 1 \\ μ = 1 + \frac{1}{2} \end{array}$
Hence, the refractive index of the liquid-
$μ = \frac{3}{2}$

AP EAMCET (23.04.2019) Shift-I

Ray Optics

282032 For concave mirror, if the object is just beyond the focal length, the image formed is

1 real, inverted and magnified

2 real, inverted and diminshed

3 real, erect and magnified

4 real, erect and diminished

МНТ-СЕТ 2019

Ray Optics

282033 An object is placed at a distance of $40 cm$ from a concave mirror of focal length $15 cm$ . If the object is displaced through a distance of $20 cm$ towards the mirror, the displacement of the image will be

1

30 cm

towards the mirror

2

36 cm

away from the mirror

3

30 cm

away from the mirror

4

36 cm

toward the mirror

Explanation:

B: Case - I
Given, object distance $(u) = - 40 cm$
Focal length $(f) = - 15 cm$
From mirror formula
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{- 1}{15} = \frac{- 1}{40} + \frac{1}{v} \\ \frac{1}{V} = \frac{1}{40} - \frac{1}{15} \\ \frac{1}{v} = \frac{3 - 8}{120} \\ \frac{1}{V} = \frac{- 5}{120} \\ v = - 24 cm \end{array}$
Hence, image is formed in front of mirror at a distance $24 cm$ .
Case - II
Focus remains same
$f = - 15 cm$
Object new location from the mirror
$\begin{array}{r} u^{'} = u - 20 \\ u^{'} = 40 - 20 \\ u^{'} = 20 cm \end{array}$
Hence, object new position is $- 20 cm$ from the mirror formula
$\begin{aligned} \frac{1}{f} & = \frac{1}{u^{'}} + \frac{1}{v^{'}} \\ \frac{1}{v^{'}} & = \frac{1}{20} - \frac{1}{15} \\ \frac{1}{v^{'}} & = \frac{3 - 4}{60} \\ v^{'} & = - 60 cm \end{aligned}$
Hence, image move by
$\begin{array}{r} = v^{'} - v \\ = - 60 - (- 24) \\ = - 60 + 24 \\ = - 36 cm \end{array}$
Hence, image move by $36 cm$ away from the mirror.

NEET- 2018

Ray Optics

282034 Parallel rays of light entering a convex lens always converge at

1 Centre of curvature

2 The principle focus

3 Optical centre

4 Focal plane

SRM JEE-2018

Ray Optics

282031 When an object is more along the principle axis of a concave mirror placed in air, the image coincides with the object if the object is $50 cm$ from the mirror. If the mirror is place at the depth of $20 cm$ in transparent medium, the image coincided with the object when the object is $40 cm$ from the mirror. The refractive index of the liquid is

1

\frac{5}{4}

2

\frac{4}{3}

3

\frac{3}{2}

4

\frac{5}{3}

Explanation:

C: Given,
When liquid is not present $u = - 50 cm$ and $v = - 50 cm$
Then, for focal length -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{50} - \frac{1}{50} \\ \frac{1}{f} = - \frac{2}{50} \\ f = - 25 cm \end{array}$
Lateral shift in liquid $= 20 cm$
Ray travel in liquid $= 40 cm$
We know,
Lateral shift $=$ Ray travel in liquid $(μ - 1)$
$\begin{array}{r} 20 = 40 (μ - 1) \\ \frac{1}{2} = μ - 1 \\ μ = 1 + \frac{1}{2} \end{array}$
Hence, the refractive index of the liquid-
$μ = \frac{3}{2}$

AP EAMCET (23.04.2019) Shift-I

Ray Optics

282032 For concave mirror, if the object is just beyond the focal length, the image formed is

1 real, inverted and magnified

2 real, inverted and diminshed

3 real, erect and magnified

4 real, erect and diminished

МНТ-СЕТ 2019

Ray Optics

282033 An object is placed at a distance of $40 cm$ from a concave mirror of focal length $15 cm$ . If the object is displaced through a distance of $20 cm$ towards the mirror, the displacement of the image will be

1

30 cm

towards the mirror

2

36 cm

away from the mirror

3

30 cm

away from the mirror

4

36 cm

toward the mirror

Explanation:

B: Case - I
Given, object distance $(u) = - 40 cm$
Focal length $(f) = - 15 cm$
From mirror formula
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{- 1}{15} = \frac{- 1}{40} + \frac{1}{v} \\ \frac{1}{V} = \frac{1}{40} - \frac{1}{15} \\ \frac{1}{v} = \frac{3 - 8}{120} \\ \frac{1}{V} = \frac{- 5}{120} \\ v = - 24 cm \end{array}$
Hence, image is formed in front of mirror at a distance $24 cm$ .
Case - II
Focus remains same
$f = - 15 cm$
Object new location from the mirror
$\begin{array}{r} u^{'} = u - 20 \\ u^{'} = 40 - 20 \\ u^{'} = 20 cm \end{array}$
Hence, object new position is $- 20 cm$ from the mirror formula
$\begin{aligned} \frac{1}{f} & = \frac{1}{u^{'}} + \frac{1}{v^{'}} \\ \frac{1}{v^{'}} & = \frac{1}{20} - \frac{1}{15} \\ \frac{1}{v^{'}} & = \frac{3 - 4}{60} \\ v^{'} & = - 60 cm \end{aligned}$
Hence, image move by
$\begin{array}{r} = v^{'} - v \\ = - 60 - (- 24) \\ = - 60 + 24 \\ = - 36 cm \end{array}$
Hence, image move by $36 cm$ away from the mirror.

NEET- 2018

Ray Optics

282034 Parallel rays of light entering a convex lens always converge at

1 Centre of curvature

2 The principle focus

3 Optical centre

4 Focal plane

SRM JEE-2018

Ray Optics

282031 When an object is more along the principle axis of a concave mirror placed in air, the image coincides with the object if the object is $50 cm$ from the mirror. If the mirror is place at the depth of $20 cm$ in transparent medium, the image coincided with the object when the object is $40 cm$ from the mirror. The refractive index of the liquid is

1

\frac{5}{4}

2

\frac{4}{3}

3

\frac{3}{2}

4

\frac{5}{3}

Explanation:

C: Given,
When liquid is not present $u = - 50 cm$ and $v = - 50 cm$
Then, for focal length -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{50} - \frac{1}{50} \\ \frac{1}{f} = - \frac{2}{50} \\ f = - 25 cm \end{array}$
Lateral shift in liquid $= 20 cm$
Ray travel in liquid $= 40 cm$
We know,
Lateral shift $=$ Ray travel in liquid $(μ - 1)$
$\begin{array}{r} 20 = 40 (μ - 1) \\ \frac{1}{2} = μ - 1 \\ μ = 1 + \frac{1}{2} \end{array}$
Hence, the refractive index of the liquid-
$μ = \frac{3}{2}$

AP EAMCET (23.04.2019) Shift-I

Ray Optics

282032 For concave mirror, if the object is just beyond the focal length, the image formed is

1 real, inverted and magnified

2 real, inverted and diminshed

3 real, erect and magnified

4 real, erect and diminished

МНТ-СЕТ 2019

Ray Optics

282033 An object is placed at a distance of $40 cm$ from a concave mirror of focal length $15 cm$ . If the object is displaced through a distance of $20 cm$ towards the mirror, the displacement of the image will be

1

30 cm

towards the mirror

2

36 cm

away from the mirror

3

30 cm

away from the mirror

4

36 cm

toward the mirror

Explanation:

B: Case - I
Given, object distance $(u) = - 40 cm$
Focal length $(f) = - 15 cm$
From mirror formula
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{- 1}{15} = \frac{- 1}{40} + \frac{1}{v} \\ \frac{1}{V} = \frac{1}{40} - \frac{1}{15} \\ \frac{1}{v} = \frac{3 - 8}{120} \\ \frac{1}{V} = \frac{- 5}{120} \\ v = - 24 cm \end{array}$
Hence, image is formed in front of mirror at a distance $24 cm$ .
Case - II
Focus remains same
$f = - 15 cm$
Object new location from the mirror
$\begin{array}{r} u^{'} = u - 20 \\ u^{'} = 40 - 20 \\ u^{'} = 20 cm \end{array}$
Hence, object new position is $- 20 cm$ from the mirror formula
$\begin{aligned} \frac{1}{f} & = \frac{1}{u^{'}} + \frac{1}{v^{'}} \\ \frac{1}{v^{'}} & = \frac{1}{20} - \frac{1}{15} \\ \frac{1}{v^{'}} & = \frac{3 - 4}{60} \\ v^{'} & = - 60 cm \end{aligned}$
Hence, image move by
$\begin{array}{r} = v^{'} - v \\ = - 60 - (- 24) \\ = - 60 + 24 \\ = - 36 cm \end{array}$
Hence, image move by $36 cm$ away from the mirror.

NEET- 2018

Ray Optics

282034 Parallel rays of light entering a convex lens always converge at

1 Centre of curvature

2 The principle focus

3 Optical centre

4 Focal plane

SRM JEE-2018