17
(a) Write the name of the lens and position of the object in front of it so that a virtual and magnified image is formed.
(b) An object is placed perpendicular to the principal axis of a convex lens of focal length $$\begin{aligned}& 20 \\ & \text{cm}\end{aligned}$$. If object is located at $$\begin{aligned}& 30 \\ & \text{cm}\end{aligned}$$ from this lens, using lens formula find out the position and nature of the image formed.
(b) An object is placed perpendicular to the principal axis of a convex lens of focal length $$\begin{aligned}& 20 \\ & \text{cm}\end{aligned}$$. If object is located at $$\begin{aligned}& 30 \\ & \text{cm}\end{aligned}$$ from this lens, using lens formula find out the position and nature of the image formed.
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(a)
* Convex lens (½ Mark)
* When object is placed between focus and optical centre (½ Mark)
(b) $f = + 20 \text{ cm}$
$u = -30 \text{ cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-30}$
$\frac{1}{v} = \frac{1}{60}$ (½ Mark)
$v = 60 \text{ cm}$
The image is formed at a distance of 60 cm on the other side of the optical centre.
* Nature of image – Real, inverted (1 Mark)
* Convex lens (½ Mark)
* When object is placed between focus and optical centre (½ Mark)
(b) $f = + 20 \text{ cm}$
$u = -30 \text{ cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-30}$
$\frac{1}{v} = \frac{1}{60}$ (½ Mark)
$v = 60 \text{ cm}$
The image is formed at a distance of 60 cm on the other side of the optical centre.
* Nature of image – Real, inverted (1 Mark)