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A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.?
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To find the distance at which the object should be placed from the lens, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
Given:
f = -15 cm (negative sign indicates a concave lens)
v = -10 cm (negative sign indicates a real image formed on the opposite side of the lens)
u = ?
Let's plug in the values into the lens formula and solve for u:
1/-15 = 1/-10 - 1/u
Multiplying through by -15u, we get:
u = -15(-10) / (-15 - 10)
u = 150 / 25
u = 6 cm
Therefore, the object should be placed 6 cm from the lens in order to form an image 10 cm from the lens.
Now, let's determine the nature and size of the image.
The image formed by a concave lens is always virtual, erect, and diminished in size. This means that the image will be upright, smaller than the object, and located on the same side as the object.
Since the object is 2 cm tall, the size of the image will be smaller than 2 cm.
Using the magnification formula:
magnification = height of image / height of object
We can find the magnification of the lens:
magnification = v/u = -10/6 = -5/3
The negative sign indicates that the image is inverted, but since the image is virtual, it is actually upright. Therefore, the magnification in this case would be positive, which means the image is magnified.
To find the height of the image, we can use the magnification formula:
magnification = height of image / height of object
Rearranging the formula, we get:
height of image = magnification * height of object
height of image = (5/3) * 2 cm
height of image = 10/3 cm
Therefore, the nature of the image is virtual, erect, and magnified, and its height is 10/3 cm.
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
Given:
f = -15 cm (negative sign indicates a concave lens)
v = -10 cm (negative sign indicates a real image formed on the opposite side of the lens)
u = ?
Let's plug in the values into the lens formula and solve for u:
1/-15 = 1/-10 - 1/u
Multiplying through by -15u, we get:
u = -15(-10) / (-15 - 10)
u = 150 / 25
u = 6 cm
Therefore, the object should be placed 6 cm from the lens in order to form an image 10 cm from the lens.
Now, let's determine the nature and size of the image.
The image formed by a concave lens is always virtual, erect, and diminished in size. This means that the image will be upright, smaller than the object, and located on the same side as the object.
Since the object is 2 cm tall, the size of the image will be smaller than 2 cm.
Using the magnification formula:
magnification = height of image / height of object
We can find the magnification of the lens:
magnification = v/u = -10/6 = -5/3
The negative sign indicates that the image is inverted, but since the image is virtual, it is actually upright. Therefore, the magnification in this case would be positive, which means the image is magnified.
To find the height of the image, we can use the magnification formula:
magnification = height of image / height of object
Rearranging the formula, we get:
height of image = magnification * height of object
height of image = (5/3) * 2 cm
height of image = 10/3 cm
Therefore, the nature of the image is virtual, erect, and magnified, and its height is 10/3 cm.
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A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.?
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A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.?.
A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2 cm tall object is placed perpendicular to the principal axis of a concave lens of focal length 15cm.At what distance from the lens should the object be placed so that it forms an image 10cm from the lens?Also find the nature and size of the image.?.
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