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(i) Define the focus of a convex lens.
(ii) An object placed on a meter scale at 8 $\text{cm}$ mark, was focused on a white screen placed at 92 $\text{cm}$ mark, by using 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 object if the image forms at 71 $\text{cm}$ mark on meter scale.
(III) State the nature of image formed if the object is kept on 36 $\text{cm}$ mark of meter scale.
(ii) An object placed on a meter scale at 8 $\text{cm}$ mark, was focused on a white screen placed at 92 $\text{cm}$ mark, by using 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 object if the image forms at 71 $\text{cm}$ mark on meter scale.
(III) State the nature of image formed if the object is kept on 36 $\text{cm}$ mark of meter scale.
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It is a point on principal axis where light rays incident parallel to principal axis meet (converge) after refraction through a convex lens. (1 Mark)
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{42} - (\frac{1}{-42}) = \frac{1}{f}$ (½ Mark)
$f = 21 \text{ cm}$ (½ Mark)
Alternate answer for (I):
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
Since image distance is same as object distance. So, object is placed at $2F_1$. (½ Mark)
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (½ Mark)
$v = (71 - 50) \text{ cm} = 21 \text{ cm}$ (½ Mark)
Since, image is formed at focus $F_2$. So, object is at infinity. (1 Mark)
$u = - (50 - 36) \text{ cm} = -14 \text{ cm}$ (½ Mark)
Since, object is kept between focus $F_1$ and optical centre. So, nature of image formed will be virtual and erect. (1 Mark)
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{42} - (\frac{1}{-42}) = \frac{1}{f}$ (½ Mark)
$f = 21 \text{ cm}$ (½ Mark)
Alternate answer for (I):
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
Since image distance is same as object distance. So, object is placed at $2F_1$. (½ Mark)
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (½ Mark)
$v = (71 - 50) \text{ cm} = 21 \text{ cm}$ (½ Mark)
Since, image is formed at focus $F_2$. So, object is at infinity. (1 Mark)
$u = - (50 - 36) \text{ cm} = -14 \text{ cm}$ (½ Mark)
Since, object is kept between focus $F_1$ and optical centre. So, nature of image formed will be virtual and erect. (1 Mark)