
A convex lens is placed somewhere in between an object and screen. The distance between an object and screen is $48cm$. If the numerical value of magnification produced by the lens is 3 , Focal length of lens is:
A. $16cm$
B. $4.5cm$
C. $12cm$
D. $9cm$
Answer
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Hint: This type of lens is thicker at the center and thinner at the edges. If those surfaces are bent outwards, the lens is called a biconvex lens or simply convex lens. These types of lenses can converge a beam of light coming from outside and focus it to a point on the other side.
Step by step answer: The distance between an object and screen = $v = 48cm$
Magnification produced by the lens = $3$
The focal length of a thin convex lens can be easily measured by using it to form an image of a distant light source on a screen. The lens is moved until a sharp image is formed on the screen.
The central length of a focal point decides the amplification at which it pictures removed articles. It is equivalent to the separation between the picture plane and a pinhole that pictures inaccessible items a similar size as the focal point being referred to.
The focal length of such a focal point is viewed as where the spreading light emissions would meet before the focal point if the focal point were not there.
Focal length of convex lens is given by
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$………………………………………………………………………………….... (I)
Where $f$= focal length of the convex lens
Magnification $M$ of convex lens is $3$
$M = \dfrac{v}{u}$
$ 3 = \dfrac{{48}}{u}$
$\Rightarrow u = \dfrac{{48}}{3}$
$\therefore u = 16cm$
Put value $v = 48cm$ and $u = 16cm$ in equation (I):
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{48}} + \dfrac{1}{{16}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{{16 + 48}}{{16 \times 48}}$
$\Rightarrow f = \dfrac{{16 \times 48}}{{48 + 16}}$
$\Rightarrow f = \dfrac{{768}}{{64}}$
$\therefore f = 12cm$
Hence, option (C) is the correct answer.
Note: This type of lens is thicker at the centre and thinner at the edges. If those surfaces are bent outwards, the lens is called a biconvex lens or simply convex lens.Most of the students make mistakes while taking signs of the image or object distance, because of improper sign convention taken.
Step by step answer: The distance between an object and screen = $v = 48cm$
Magnification produced by the lens = $3$
The focal length of a thin convex lens can be easily measured by using it to form an image of a distant light source on a screen. The lens is moved until a sharp image is formed on the screen.
The central length of a focal point decides the amplification at which it pictures removed articles. It is equivalent to the separation between the picture plane and a pinhole that pictures inaccessible items a similar size as the focal point being referred to.
The focal length of such a focal point is viewed as where the spreading light emissions would meet before the focal point if the focal point were not there.
Focal length of convex lens is given by
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$………………………………………………………………………………….... (I)
Where $f$= focal length of the convex lens
Magnification $M$ of convex lens is $3$
$M = \dfrac{v}{u}$
$ 3 = \dfrac{{48}}{u}$
$\Rightarrow u = \dfrac{{48}}{3}$
$\therefore u = 16cm$
Put value $v = 48cm$ and $u = 16cm$ in equation (I):
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{48}} + \dfrac{1}{{16}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{{16 + 48}}{{16 \times 48}}$
$\Rightarrow f = \dfrac{{16 \times 48}}{{48 + 16}}$
$\Rightarrow f = \dfrac{{768}}{{64}}$
$\therefore f = 12cm$
Hence, option (C) is the correct answer.
Note: This type of lens is thicker at the centre and thinner at the edges. If those surfaces are bent outwards, the lens is called a biconvex lens or simply convex lens.Most of the students make mistakes while taking signs of the image or object distance, because of improper sign convention taken.
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