
A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens.
A) must be less than 10 cm
B) must be greater than 200 cm
C) must not be greater than 200cm
D) must not be less than 10 cm
Answer
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Hint: According to the question we have to write given values, substituting those given values in the lens formula with proper sign conversion from this we get u value. There we can analyse the focal length of lense from which we choose the correct answer.
Formula used:
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Complete answer:
Let d is the distance between source and screen
f is the focal length
u is the distance between source and lense
so, we have v=d-u for getting image on the screen
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Substituting given values we get
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{d - u}} - \dfrac{1}{{ - u}}$
$ \Rightarrow u = \dfrac{{d \pm \sqrt {{d^2} - 4df} }}{2}$
$ \Rightarrow u = \dfrac{{d \pm \sqrt {d\left( {d - 4f} \right)} }}{2}$
From the above equation , we can see that to get real values of u and v. then we must have d> 4f.
When we have d=4f we get only one value of $u = \dfrac{d}{2}$
When we have d>4f, we get two values of u for which we can get a real image on screen.
In this problem we are not getting an image on screen it means d<4f i.e, 4f>d, f>10cm.
Therefore, the answer is f must be greater than 10 cm.
Hence, the correct option is D.
Additional formula:
Lens formula is applicable for both convex as well as concave lenses. Because these lenses have negligible thickness. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. This lens formula is applicable to all situations with appropriate sign conventions.
Note:
Students may forget the sign convention when we put the values in the lense formula where we have an algebraic form of the equation. We have to clearly hold on to the use of lense formula. We have another way to choose the correct option by therioticaly just to find u value.
Formula used:
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Complete answer:
Let d is the distance between source and screen
f is the focal length
u is the distance between source and lense
so, we have v=d-u for getting image on the screen
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Substituting given values we get
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{d - u}} - \dfrac{1}{{ - u}}$
$ \Rightarrow u = \dfrac{{d \pm \sqrt {{d^2} - 4df} }}{2}$
$ \Rightarrow u = \dfrac{{d \pm \sqrt {d\left( {d - 4f} \right)} }}{2}$
From the above equation , we can see that to get real values of u and v. then we must have d> 4f.
When we have d=4f we get only one value of $u = \dfrac{d}{2}$
When we have d>4f, we get two values of u for which we can get a real image on screen.
In this problem we are not getting an image on screen it means d<4f i.e, 4f>d, f>10cm.
Therefore, the answer is f must be greater than 10 cm.
Hence, the correct option is D.
Additional formula:
Lens formula is applicable for both convex as well as concave lenses. Because these lenses have negligible thickness. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. This lens formula is applicable to all situations with appropriate sign conventions.
Note:
Students may forget the sign convention when we put the values in the lense formula where we have an algebraic form of the equation. We have to clearly hold on to the use of lense formula. We have another way to choose the correct option by therioticaly just to find u value.
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