
The power of a convex lens is \[ + 4.0\,D\] . At what distance should the object from the lens be placed to obtain its real and inverted image of the same size on the screen?
Answer
479.7k+ views
Hint: As we already have the power of a convex lens, we will first show the relation between power and the focal length of the given lens because focal length is necessary to apply the lens maker formula. Then, we have focal length and according to the question, the magnification of the lens is 1. And due to this, image and object distances are the same. Now we can easily find the object distance by applying the lens maker formula.
Complete step by step answer:
Given the power of a convex lens, $P = + 4D$. As we all know, power is inversely proportional to the focal length of the given lens:
$\text{Power} = \dfrac{1}{\text{Focal length}}$
We can also write the upper formula in terms of focal length as:
$\text{Focal length} = \dfrac{1}{\text{Power}}$
Now, we will solve for the focal length, as the value of power is already given:
$f = \dfrac{1}{4} \\
\Rightarrow f = 0.25\,m \\ $
Now, as per the Question, Size of the Image is same as that of Size of Object.That means, the linear magnification is 1.Also,
$m = \dfrac{v}{u}$
where, $v$ is the image distance and $u$ is the object distance. Since, magnification or $m$ is 1 , then the $v$ and $u$ are also equal to each other:
$ \Rightarrow v = u$
So, now we will using Lens formula:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
where, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance.
As we already conclude the value of focal length, and we have also $v = u$ , so we will put the values in the lens formula:
$\dfrac{1}{{0.25}} = \dfrac{1}{u} - \dfrac{1}{{ - u}}$..........(Object distance will be negative)
$\Rightarrow \dfrac{{100}}{{25}} = \dfrac{2}{u} \\
\Rightarrow 4 = \dfrac{2}{u} \\
\Rightarrow u = 0.5m \\ $
$\therefore u = 50\,cm$
Therefore, the object will be placed at a distance of $50\,cm$ from the lens.
Note: Converging (convex) and divergent lenses are the two types of lenses (concave). The focal length, radii of curvature of the curved surfaces, and the refractive index of the transparent material are all calculated using the lens maker's formula. The formula is used to create lenses with the focal lengths that are desired.
Complete step by step answer:
Given the power of a convex lens, $P = + 4D$. As we all know, power is inversely proportional to the focal length of the given lens:
$\text{Power} = \dfrac{1}{\text{Focal length}}$
We can also write the upper formula in terms of focal length as:
$\text{Focal length} = \dfrac{1}{\text{Power}}$
Now, we will solve for the focal length, as the value of power is already given:
$f = \dfrac{1}{4} \\
\Rightarrow f = 0.25\,m \\ $
Now, as per the Question, Size of the Image is same as that of Size of Object.That means, the linear magnification is 1.Also,
$m = \dfrac{v}{u}$
where, $v$ is the image distance and $u$ is the object distance. Since, magnification or $m$ is 1 , then the $v$ and $u$ are also equal to each other:
$ \Rightarrow v = u$
So, now we will using Lens formula:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
where, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance.
As we already conclude the value of focal length, and we have also $v = u$ , so we will put the values in the lens formula:
$\dfrac{1}{{0.25}} = \dfrac{1}{u} - \dfrac{1}{{ - u}}$..........(Object distance will be negative)
$\Rightarrow \dfrac{{100}}{{25}} = \dfrac{2}{u} \\
\Rightarrow 4 = \dfrac{2}{u} \\
\Rightarrow u = 0.5m \\ $
$\therefore u = 50\,cm$
Therefore, the object will be placed at a distance of $50\,cm$ from the lens.
Note: Converging (convex) and divergent lenses are the two types of lenses (concave). The focal length, radii of curvature of the curved surfaces, and the refractive index of the transparent material are all calculated using the lens maker's formula. The formula is used to create lenses with the focal lengths that are desired.
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