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The focal length of a convex lens is 20 cm At what distance from the lens should an object be placed so that a two times larger and real image may be obtained?
ans=30 but how?
Most Upvoted Answer
The focal length of a convex lens is 20 cm At what distance from the l...
We know that m=v/u=h'/h given:h'=2h covex lens =)1/v - 1/u = 1/f let v be 2x and u be -x =)1/2x +1/x = -1/20 =)x + 2x/2x square = -1/20 =) 3x/2x square =-1/20 =)60 x =-2x square =)60/-2=x square/x =)-30=x(because object is always placed on left side) =)u=30
Community Answer
The focal length of a convex lens is 20 cm At what distance from the l...
Introduction:
The focal length of a convex lens is 20 cm. We need to find the distance from the lens at which an object should be placed to obtain a two times larger and real image.

Given:
Focal length of convex lens (f) = 20 cm

Formula:
The formula relating the object distance (u), image distance (v), and focal length (f) of a lens is:

1/f = 1/v - 1/u

Where,
f = focal length of the lens
v = image distance
u = object distance

Step 1: Find the image distance for the given condition:
We are given that the image formed is two times larger than the object. This implies the magnification (m) is 2.

The formula for magnification is:
m = -v/u

Given m = 2, we can rearrange the formula to find the image distance (v):
v = -2u

Step 2: Substitute the values into the lens formula:
Now, we substitute the values of f and v into the lens formula:
1/20 = 1/(-2u) - 1/u

Step 3: Solve the equation:
To solve this equation, we can find a common denominator and simplify:
1/20 = -1/2u - 2/2u
1/20 = -3/2u
Cross-multiplying:
2u = -60
u = -30 cm

Step 4: Determine the positive distance:
The object distance (u) is calculated as -30 cm, but distance cannot be negative. Therefore, we take the positive value of u:
u = 30 cm

Conclusion:
So, the object should be placed at a distance of 30 cm from the lens in order to obtain a two times larger and real image.
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The focal length of a convex lens is 20 cm At what distance from the lens should an object be placed so that a two times larger and real image may be obtained?ans=30 but how?
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