Question

The relationship between the distance of object from the lens (u), distance of image from the lens (v) and the focal length (j) of the lens is called lens formula. It can be written as1f=1v−1u
The size of image formed by a lens depends on the position of the object from the lens. A lens of short focal length has more power whereas a lens of long focal length has less power. When the lens is convex, the power is positive and for concave lens, the power is negative.
The magnification produced by a lens is the ratio of height of image to the height of object as the size of the image relative to the object is given by linear magnification (m).
When, m is negative, image formed is real and when m is positive, image formed is virtual. If m < 1, size of image is smaller than the object. If m > 1, size of image is larger than the object.
(i) An object 4 cm in height is placed at a distance of 10 cm from a convex lens of focal length 20 cm. The position of image is
(a) - 20 cm (b) 20 cm
(c) -10 cm (d) 10 cm
(ii) In the above question, the size of image is
(a) 16 cm (b) 8 cm
(c) 4 cm (d) 2 cm
(iii) An object is ,placed 50 cm from a concave lens and produces a virtual image at a distance of 10 cm in front of lens. The focal length of lens is
(a) - 25 cm (b) -12.5 cm
(c) 12.5 cm (d) 10 cm
(iv) A convex lens forms an image of magnification -2 of the height of image is 6 cm, the height of object is
(a) 6 cm (b) 4 cm
(c) 3 cm (d) 2 cm
(v) A concave lens of focal length 5 ern, the power of lens is
(a) 20D (b) -20D
(c) 90D (d) -5 D

Answer 1

Answer:

(i) a

(ii) b

(iii) d

(iv) c

Explanation:

Answer 2

The Image distance in the convex lens will be (a) -20cm.

The image height for the convex lens will be (b) 8 cm.

The focal length of the concave lens is (b) -12.5 cm.

The height of the object will be (c) 3 cm.

The power of the lens of focal length will be (b) -20D.

Explanation 1

We have been given in a convex lens that

Object distance $(u) = -10 cm$   $;$ $and$

Focal length $(f)= +20 cm$

According to lens law, we have

$\frac{1}{v}- \frac{1}{u}= \frac{1}{f}$   $;$ $or$

$v=\frac{uf}{u+f}$

Substituting the given values, we get

$v=\frac{(-10)(20)}{(-10)+(20)}$

Hence, image distance $(v)$ will be $(a)$ $-20cm.$

Explanation 2

Also, we have been given in the question that

Object height $(h)=+4cm$

We know,

Magnification of an object is given by,

$m=\frac{v}{u}=\frac{h'}{h}$  

Substituting the known values in the given equation, we get

$\frac{-20}{-10}=\frac{h'}{4}$  $;$  $or$

Hence, Image height $(h')$ will be $(b)$ $8 cm.$

Explanation 3

According to the question, we have

Object distance $(u)=-50cm$

Image distance $(v)= -10cm$

According to Lens law,

$\frac{1}{v}- \frac{1}{u}= \frac{1}{f}$  $;$ $or$

$f=\frac{uv}{u-v}$  

Substituting the given values in the equation, we get

$f=\frac{(-50)(-10)}{(-50)-(-10)}$

Hence, the focal length $(f)$ of the given concave lens is $(b)-12.5cm.$

Explanation 4

Given that,

Magnification $(m)=-2$

Image height $(h')=6cm$

Magnification for lens is given by

$m=\frac{h'}{h}$  $;$  $or$

$h=\frac{h'}{m}$

Substituting the given values in the equation, we get

$h=\frac{6}{2}$

Hence, object height $(h)$ is $(c)3 cm.$

Explanation 5

Given,

Focal length of the concave lens is always negative hence, here

$f=-0.05 cm$

We know,

Power of lens $P=\frac{1}{f}$

Substituting the given values in the equation, we get

$P=\frac{1}{-0.05}$

Hence, the Power of the concave lens will be $(b)-20D.$