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A 2cm tall object is placed perpendicular to the principal axis of 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 image formed.?
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Given:
- Height of object (h) = 2 cm
- Focal length of concave lens (f) = -15 cm (negative sign indicates that the lens is concave)
- Distance of image (v) = 10 cm
To find:
- Distance of object (u)
- Nature and size of the image formed
Solution:
Step 1: Identify the given and required quantities:
Given:
- Height of object (h) = 2 cm
- Focal length of concave lens (f) = -15 cm
- Distance of image (v) = 10 cm
Required:
- Distance of object (u)
- Nature and size of the image formed
Step 2: Apply the lens formula:
The lens formula is given by:
1/f = 1/v - 1/u
Substituting the given values, we get:
1/-15 = 1/10 - 1/u
Simplifying the equation, we get:
-1/15 = 1/10 - 1/u
Step 3: Solve for u:
To find the value of u, we need to solve the equation obtained in step 2.
Multiplying both sides of the equation by 150u, we get:
-10u = 15u - 150
Combining like terms, we get:
-10u - 15u = -150
-25u = -150
Dividing both sides of the equation by -25, we get:
u = -150/-25
Simplifying the equation, we get:
u = 6 cm
Therefore, the object should be placed at a distance of 6 cm from the lens.
Step 4: Determine the nature and size of the image:
To determine the nature and size of the image formed, we can use the magnification formula:
Magnification (m) = -v/u
Substituting the given values, we get:
m = -10/6
Simplifying the equation, we get:
m = -5/3
Since the magnification is negative, the image formed will be inverted.
To find the size of the image, we can use the magnification formula:
m = height of image/height of object
Substituting the given values, we get:
-5/3 = height of image/2
Simplifying the equation, we get:
height of image = (-5/3) * 2
height of image = -10/3 cm
Therefore, the image formed is inverted and its size is -10/3 cm.
Conclusion:
The object should be placed at a distance of 6 cm from the concave lens. The image formed will be inverted and its size will be -10/3 cm.
- Height of object (h) = 2 cm
- Focal length of concave lens (f) = -15 cm (negative sign indicates that the lens is concave)
- Distance of image (v) = 10 cm
To find:
- Distance of object (u)
- Nature and size of the image formed
Solution:
Step 1: Identify the given and required quantities:
Given:
- Height of object (h) = 2 cm
- Focal length of concave lens (f) = -15 cm
- Distance of image (v) = 10 cm
Required:
- Distance of object (u)
- Nature and size of the image formed
Step 2: Apply the lens formula:
The lens formula is given by:
1/f = 1/v - 1/u
Substituting the given values, we get:
1/-15 = 1/10 - 1/u
Simplifying the equation, we get:
-1/15 = 1/10 - 1/u
Step 3: Solve for u:
To find the value of u, we need to solve the equation obtained in step 2.
Multiplying both sides of the equation by 150u, we get:
-10u = 15u - 150
Combining like terms, we get:
-10u - 15u = -150
-25u = -150
Dividing both sides of the equation by -25, we get:
u = -150/-25
Simplifying the equation, we get:
u = 6 cm
Therefore, the object should be placed at a distance of 6 cm from the lens.
Step 4: Determine the nature and size of the image:
To determine the nature and size of the image formed, we can use the magnification formula:
Magnification (m) = -v/u
Substituting the given values, we get:
m = -10/6
Simplifying the equation, we get:
m = -5/3
Since the magnification is negative, the image formed will be inverted.
To find the size of the image, we can use the magnification formula:
m = height of image/height of object
Substituting the given values, we get:
-5/3 = height of image/2
Simplifying the equation, we get:
height of image = (-5/3) * 2
height of image = -10/3 cm
Therefore, the image formed is inverted and its size is -10/3 cm.
Conclusion:
The object should be placed at a distance of 6 cm from the concave lens. The image formed will be inverted and its size will be -10/3 cm.
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A 2cm tall object is placed perpendicular to the principal axis of 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 image formed.?
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A 2cm tall object is placed perpendicular to the principal axis of 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 image formed.? 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 2cm tall object is placed perpendicular to the principal axis of 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 image formed.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2cm tall object is placed perpendicular to the principal axis of 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 image formed.?.
A 2cm tall object is placed perpendicular to the principal axis of 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 image formed.? 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 2cm tall object is placed perpendicular to the principal axis of 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 image formed.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2cm tall object is placed perpendicular to the principal axis of 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 image formed.?.
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