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(i) Define the optical centre of a lens.
(ii) The light from an object placed on a meter scale at $8\ \text{cm}$ mark, was focused on a white screen placed at $92\ \text{cm}$ mark, with the help of a converging lens placed on the scale at $50\ \text{cm}$ mark :
(I) Find the focal length of converging lens.
(II) Find the position of image if the object is shifted towards the lens and kept at $29\ \text{cm}$ mark.
(III) What will be the nature of the image formed if the object is further shifted towards the lens ?
(ii) The light from an object placed on a meter scale at $8\ \text{cm}$ mark, was focused on a white screen placed at $92\ \text{cm}$ mark, with the help of a converging lens placed on the scale at $50\ \text{cm}$ mark :
(I) Find the focal length of converging lens.
(II) Find the position of image if the object is shifted towards the lens and kept at $29\ \text{cm}$ mark.
(III) What will be the nature of the image formed if the object is further shifted towards the lens ?
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The central point of a lens through which a ray of light passes without any deviation. (1 Mark)
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{42} - \frac{1}{-42} = \frac{1}{f}$
$f = 21 \text{ cm}$ (1 Mark)
Alternate answer for (I):
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
Since object distance is same as image distance. So, object is placed at $2F_1$.
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (1 Mark)
$u = -(50 - 29) = -21 \text{ cm}$
As now object is kept at focus $F_1$, image will be formed at infinity (1 Mark)
Now, object is kept between Optical centre and focus $F_1$,
Nature of image is virtual and erect. (1 Mark)
The central point of a lens through which a ray of light passes without any deviation. (1 Mark)
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{42} - \frac{1}{-42} = \frac{1}{f}$
$f = 21 \text{ cm}$ (1 Mark)
Alternate answer for (I):
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
Since object distance is same as image distance. So, object is placed at $2F_1$.
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (1 Mark)
$u = -(50 - 29) = -21 \text{ cm}$
As now object is kept at focus $F_1$, image will be formed at infinity (1 Mark)
Now, object is kept between Optical centre and focus $F_1$,
Nature of image is virtual and erect. (1 Mark)