A 2cm tall object is placed perpendicular to the principal axis of con...
A 2cm tall object is placed perpendicular to the principal axis of con...
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.
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