For distant vision = −0.181 m, for near vision = 0.667 m The power P of a lens of focal length f is given by the relation P = 1/f(in metres) (i) Power of the lens used for correcting distant vision = −5.5 D Focal length of the required lens, f =1/P F = 1/-5.5 = -0.181ₘ The focal length of the lens fRead more
For distant vision = −0.181 m, for near vision = 0.667 m
The power P of a lens of focal length f is given by the relation
P = 1/f(in metres)
(i) Power of the lens used for correcting distant vision = −5.5 D
Focal length of the required lens, f =1/P
F = 1/-5.5 = -0.181ₘ
The focal length of the lens for correcting distant vision is −0.181 m.
(ii) Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, f =1/P
F = 1/1.5 = +0.667ₘ
(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles. For more answers visit to website: htRead more
(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles.
(c) The least distance of distinct vision is the minimum distance of an object to see clear and distinct image. It is 25 cm for a young adult with normal visions. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
(c) The least distance of distinct vision is the minimum distance of an object to see clear and distinct image. It is 25 cm for a young adult with normal visions.
(d) The human eye forms the image of an object at its retina. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
(d) The human eye forms the image of an object at its retina.
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.
Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length F in metres, then P= 1/f(in metres) The S.I. unit of power of a lens is Dioptre. It is denoted by D. 1 dioptre is defined as the power of a lens of focal length 1 metre. 1 D = 1 m−¹ For more aRead more
Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of
focal length F in metres, then
P= 1/f(in metres)
The S.I. unit of power of a lens is Dioptre. It is denoted by D.
1 dioptre is defined as the power of a lens of focal length 1 metre.
1 D = 1 m−¹
Refractive index of a medium nm is related to the speed of light in that medium v by the relation: nₘ = Speed light of in air / Speed of light in the medium = c/v Where, c is the speed of light in vacuum/air The refractive index of diamond is 2.42. This suggests that the speed of light in diamond wiRead more
Refractive index of a medium nm is related to the speed of light in that medium v by the
relation:
nₘ = Speed light of in air / Speed of light in the medium = c/v
Where, c is the speed of light in vacuum/air
The refractive index of diamond is 2.42. This suggests that the speed of light in diamond
will reduce by a factor 2.42 compared to its speed in air.
Refractive index of a medium nₘ is given by, nₘ = Speed of light in vacuum / Speed of light in the medium = c/v Speed of light in vacuum, c = 3 × 10⁸ m s−¹ Refractive index of glass, n = 1.50 Speed of light in the glass, v = c/n = 3x10⁸ m s−¹ Refractive index of glass, n = 1.50 Speed of light in theRead more
Refractive index of a medium nₘ is given by,
nₘ = Speed of light in vacuum / Speed of light in the medium = c/v
Speed of light in vacuum, c = 3 × 10⁸ m s−¹
Refractive index of glass, n = 1.50
Speed of light in the glass, v = c/n = 3×10⁸ m s−¹
Refractive index of glass, n = 1.50
Speed of light in the glass, v= c/n = 3×10⁸/1.50 = 2×10⁸ ms−¹
The light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into the water will bend towards the normal. For more answerRead more
The light ray bends towards the normal.
When a ray of light travels from an optically rarer medium to an optically denser medium,
it gets bent towards the normal. Since water is optically denser than air, a ray of light
travelling from air into the water will bend towards the normal.
Magnification produced by a spherical mirror is given by the relation, m = Height of the image / Height of the Object = - Image distance / Object distance m = h₁/h₀ = -v/u Let the height of the object, ho = h Then, height of the image, hI = −3h (Image formed is real) -3h/h = -v/u v/u = 3 Object distRead more
Magnification produced by a spherical mirror is given by the relation,
m = Height of the image / Height of the Object = – Image distance / Object distance
m = h₁/h₀ = -v/u
Let the height of the object, ho = h
Then, height of the image, hI = −3h (Image formed is real)
-3h/h = -v/u
v/u = 3
Object distance, u = −10 cm v = 3 × (−10) = −30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm
in front of the given concave mirror.
A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting
For distant vision = −0.181 m, for near vision = 0.667 m The power P of a lens of focal length f is given by the relation P = 1/f(in metres) (i) Power of the lens used for correcting distant vision = −5.5 D Focal length of the required lens, f =1/P F = 1/-5.5 = -0.181ₘ The focal length of the lens fRead more
For distant vision = −0.181 m, for near vision = 0.667 m
The power P of a lens of focal length f is given by the relation
P = 1/f(in metres)
(i) Power of the lens used for correcting distant vision = −5.5 D
Focal length of the required lens, f =1/P
F = 1/-5.5 = -0.181ₘ
The focal length of the lens for correcting distant vision is −0.181 m.
(ii) Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, f =1/P
F = 1/1.5 = +0.667ₘ
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
The change in focal length of an eye lens is caused by the action of the
(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles. For more answers visit to website: htRead more
(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
The least distance of distinct vision for a young adult with normal vision is about
(c) The least distance of distinct vision is the minimum distance of an object to see clear and distinct image. It is 25 cm for a young adult with normal visions. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
(c) The least distance of distinct vision is the minimum distance of an object to see clear and distinct image. It is 25 cm for a young adult with normal visions.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
The human eye forms the image of an object at its
(d) The human eye forms the image of an object at its retina. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
(d) The human eye forms the image of an object at its retina.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens. For more answers visit to website: https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-11/
Define 1 dioptre of power of a lens.
Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length F in metres, then P= 1/f(in metres) The S.I. unit of power of a lens is Dioptre. It is denoted by D. 1 dioptre is defined as the power of a lens of focal length 1 metre. 1 D = 1 m−¹ For more aRead more
Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of
focal length F in metres, then
P= 1/f(in metres)
The S.I. unit of power of a lens is Dioptre. It is denoted by D.
1 dioptre is defined as the power of a lens of focal length 1 metre.
1 D = 1 m−¹
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/
The refractive index of diamond is 2.42. What is the meaning of this statement?
Refractive index of a medium nm is related to the speed of light in that medium v by the relation: nₘ = Speed light of in air / Speed of light in the medium = c/v Where, c is the speed of light in vacuum/air The refractive index of diamond is 2.42. This suggests that the speed of light in diamond wiRead more
Refractive index of a medium nm is related to the speed of light in that medium v by the
relation:
nₘ = Speed light of in air / Speed of light in the medium = c/v
Where, c is the speed of light in vacuum/air
The refractive index of diamond is 2.42. This suggests that the speed of light in diamond
will reduce by a factor 2.42 compared to its speed in air.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s–1.
Refractive index of a medium nₘ is given by, nₘ = Speed of light in vacuum / Speed of light in the medium = c/v Speed of light in vacuum, c = 3 × 10⁸ m s−¹ Refractive index of glass, n = 1.50 Speed of light in the glass, v = c/n = 3x10⁸ m s−¹ Refractive index of glass, n = 1.50 Speed of light in theRead more
Refractive index of a medium nₘ is given by,
nₘ = Speed of light in vacuum / Speed of light in the medium = c/v
Speed of light in vacuum, c = 3 × 10⁸ m s−¹
Refractive index of glass, n = 1.50
Speed of light in the glass, v = c/n = 3×10⁸ m s−¹
Refractive index of glass, n = 1.50
Speed of light in the glass, v= c/n = 3×10⁸/1.50 = 2×10⁸ ms−¹
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
The light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into the water will bend towards the normal. For more answerRead more
The light ray bends towards the normal.
When a ray of light travels from an optically rarer medium to an optically denser medium,
it gets bent towards the normal. Since water is optically denser than air, a ray of light
travelling from air into the water will bend towards the normal.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Magnification produced by a spherical mirror is given by the relation, m = Height of the image / Height of the Object = - Image distance / Object distance m = h₁/h₀ = -v/u Let the height of the object, ho = h Then, height of the image, hI = −3h (Image formed is real) -3h/h = -v/u v/u = 3 Object distRead more
Magnification produced by a spherical mirror is given by the relation,
m = Height of the image / Height of the Object = – Image distance / Object distance
m = h₁/h₀ = -v/u
Let the height of the object, ho = h
Then, height of the image, hI = −3h (Image formed is real)
-3h/h = -v/u
v/u = 3
Object distance, u = −10 cm v = 3 × (−10) = −30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm
in front of the given concave mirror.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/