A converging lens forms a real and inverted image of an object at a distance of 100 cm from it where should an object be placed in front of the lens , so that size of the image is twice size of object ?also calculate power
Answers
Answer:
v=100
u=v/2= -50
using lens formula
1/v- 1/u=1/f
1/100-1/-50
-50-100/-5000
f=33.33cm
p=1/f(in m)
p= 1/33.33×10^-2
p=3D
Answer:
The energy of the converging lens is -0.01 diopters.
Explanation:
We can use the lens system to locate the distance of the object from the converging lens, given the distance of the image:
1/f = 1/u + 1/v
Where,
f = focal size of the lens
u = distance of the object from the lens
v = distance of the photo from the lens
Since the photograph shaped is real and inverted, the distance of the photo is negative. Therefore, we have:
f = -v = -100 cm
To discover the distance of the object, we can use the magnification formula:
m = -v/u
Where,
m = magnification of the lens
We are given that the dimension of the photo is twice the measurement of the object. Therefore:
m = top of photo / peak of object = 2
Substituting the values of f and v, we get:
m = -f/u
2 = -(-100 cm) / u
u = 50 cm
Therefore, the object must be positioned 50 cm in the front of the lens to produce an photograph that is twice the dimension of the object.
To calculate the strength of the lens, we can use the formula:
P = 1/f
Where,
P = electricity of the lens in diopters
Substituting the price of f, we get:
P = 1/-100 cm = -0.01 diopters
Therefore, the energy of the converging lens is -0.01 diopters.
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