Focal length of concave lens (OF1), f = −15 cm Image distance, v = −10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150 U = -30 cm The negative value of u indicates that the object is placed 30 cm in front of the lens. Focal length of convexRead more
Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,
1/v – 1/u = 1/f
1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150
U = -30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens.
Focal length of convex mirror, f = +15 cm
Object distance, u = −10 cm
According to the mirror formula,
1/v – 1/u = 1/f
1/v = 1/f – 1/u = 1/15 + 1/10 = 25/150
V = 6 cm
The positive value of v indicates that the image is formed behind the mirror.
Magnification, m = – Image Distance / Object Distance = – v/u = -6/-10 = +0.6
The positive value of magnification indicates that the image formed is virtual and erect.
Focal length of concave lens (OF1), f = −15 cm Image distance, v = −10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150 U = -30 cm The negative value of u indicates that the object is placed 30 cm in front of the lens. For more answers visitRead more
Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,
1/v – 1/u = 1/f
1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150
U = -30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens.
Object distance, u = −25 cm Object height, ho = 5 cm Focal length, f = +10 cm According to the lens formula, 1/v-1/u=1/f 1/v = 1/f + 1/u = 1/10 – 1/25 =15/250 V= 250/15 = 16.66 cm The positive value of v shows that the image is formed at the other side of the lens. Magnification,m = - Image DistanceRead more
Object distance, u = −25 cm
Object height, ho = 5 cm
Focal length, f = +10 cm
According to the lens formula,
1/v-1/u=1/f
1/v = 1/f + 1/u = 1/10 – 1/25 =15/250
V= 250/15 = 16.66 cm
The positive value of v shows that the image is formed at the other side of the lens.
Magnification,m = – Image Distance / Object Distance = – v/u = -16.66 / 25 = -0.66
The negative sign shows that the image is real and formed behind the lens.
Magnification,m = Image height / Object height = H₁ / H0 = H₁=5
H₁ = m X Ho = -0.66 X 5 = -3.3 cm
The negative value of image height indicates that the image formed is inverted.
The position, size, and nature of image are shown in the following ray diagram.
Case I When the upper half of the lens is covered In this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure. Case II When the lower half ofRead more
Case I
When the upper half of the lens is covered
In this case, a ray of light coming from the object will be refracted by the lower half of
the lens. These rays meet at the other side of the lens to form the image of the given
object, as shown in the following figure.
Case II
When the lower half of the lens is covered
In this case, a ray of light coming from the object is refracted by the upper half of the
lens. These rays meet at the other side of the lens to form the image of the given object,
as shown in the following figure.
(a) Concave (b) Convex (c) Concave Explanation (a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce powerful parallel beam of light when the light source is placed at their principal focus. (b) Convex mirror is used in side/rear view mirror of a vehicle.Read more
(a) Concave
(b) Convex
(c) Concave
Explanation
(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can
produce powerful parallel beam of light when the light source is placed at their principal
focus.
(b) Convex mirror is used in side/rear view mirror of a vehicle. Convex mirrors give a
virtual, erect, and diminished image of the objects placed in front of it. Because of this,
they have a wide field of view. It enables the driver to see most of the traffic behind
him/her.
(c) Concave mirrors are convergent mirrors. That is why they are used to construct solar
furnaces. Concave mirrors converge the light incident on them at a single point known as
principal focus. Hence, they can be used to produce a large amount of heat at that point.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Focal length of concave lens (OF1), f = −15 cm Image distance, v = −10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150 U = -30 cm The negative value of u indicates that the object is placed 30 cm in front of the lens. Focal length of convexRead more
Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,
1/v – 1/u = 1/f
1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150
U = -30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens.
Focal length of convex mirror, f = +15 cm
Object distance, u = −10 cm
According to the mirror formula,
1/v – 1/u = 1/f
1/v = 1/f – 1/u = 1/15 + 1/10 = 25/150
V = 6 cm
The positive value of v indicates that the image is formed behind the mirror.
Magnification, m = – Image Distance / Object Distance = – v/u = -6/-10 = +0.6
The positive value of magnification indicates that the image formed is virtual and erect.
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A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Focal length of concave lens (OF1), f = −15 cm Image distance, v = −10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150 U = -30 cm The negative value of u indicates that the object is placed 30 cm in front of the lens. For more answers visitRead more
Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,
1/v – 1/u = 1/f
1/u = 1/v – 1/f = -1/10 – 1/(-15) = -1/10 + 1/15 = -5/150
U = -30 cm
The negative value of u indicates that the object is placed 30 cm in front of the lens.
For more answers visit to website:
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An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Object distance, u = −25 cm Object height, ho = 5 cm Focal length, f = +10 cm According to the lens formula, 1/v-1/u=1/f 1/v = 1/f + 1/u = 1/10 – 1/25 =15/250 V= 250/15 = 16.66 cm The positive value of v shows that the image is formed at the other side of the lens. Magnification,m = - Image DistanceRead more
Object distance, u = −25 cm
Object height, ho = 5 cm
Focal length, f = +10 cm
According to the lens formula,
1/v-1/u=1/f
1/v = 1/f + 1/u = 1/10 – 1/25 =15/250
V= 250/15 = 16.66 cm
The positive value of v shows that the image is formed at the other side of the lens.
Magnification,m = – Image Distance / Object Distance = – v/u = -16.66 / 25 = -0.66
The negative sign shows that the image is real and formed behind the lens.
Magnification,m = Image height / Object height = H₁ / H0 = H₁=5
H₁ = m X Ho = -0.66 X 5 = -3.3 cm
The negative value of image height indicates that the image formed is inverted.
The position, size, and nature of image are shown in the following ray diagram.
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One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Case I When the upper half of the lens is covered In this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure. Case II When the lower half ofRead more
Case I
When the upper half of the lens is covered
In this case, a ray of light coming from the object will be refracted by the lower half of
the lens. These rays meet at the other side of the lens to form the image of the given
object, as shown in the following figure.
Case II
See lessWhen the lower half of the lens is covered
In this case, a ray of light coming from the object is refracted by the upper half of the
lens. These rays meet at the other side of the lens to form the image of the given object,
as shown in the following figure.
Name the type of mirror used in the following situations.
(a) Concave (b) Convex (c) Concave Explanation (a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce powerful parallel beam of light when the light source is placed at their principal focus. (b) Convex mirror is used in side/rear view mirror of a vehicle.Read more
(a) Concave
(b) Convex
(c) Concave
Explanation
(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can
produce powerful parallel beam of light when the light source is placed at their principal
focus.
(b) Convex mirror is used in side/rear view mirror of a vehicle. Convex mirrors give a
virtual, erect, and diminished image of the objects placed in front of it. Because of this,
they have a wide field of view. It enables the driver to see most of the traffic behind
him/her.
(c) Concave mirrors are convergent mirrors. That is why they are used to construct solar
furnaces. Concave mirrors converge the light incident on them at a single point known as
principal focus. Hence, they can be used to produce a large amount of heat at that point.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-10/