Question

If an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8cm, find the position, nature and height of the image

Answer 1

U= -12cm, F=+8cm ,V=? using lens formula

1/f=1/v-1/u , 1/8=1/v-(-1/12) , 1/8-1/12=1/v , 3-2/24=1/v , 1/24=1/v , v=+24 cm , ... m=Hi/Ho=v/u , m=v/u , m=24/-12 then m is -2 and m=Hi/Ho so -2=Hi/7 then Hi is -14cm here - means inverted image form and this is real image Larger than twice Time object image

Answer 2

Given :-

• Object distance (u) = - 12 cm

• Focal length (f) = 8 cm

• Height of object (h1) = 7 cm

To find :-

• Position,nature and height of the image

Solution :-

$\red{\bigstar}\:\underbrace{\bf\green{{By\:using\:lens\:formula}}}$

$\blue{\bigstar}\:{\underline{\boxed{\bf\red{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}}}}$

$\rm\longrightarrow\:\dfrac{1}{v}-\dfrac{1}{-12}=\dfrac{1}{8}$

$\rm\longrightarrow\:\dfrac{1}{v}+\dfrac{1}{12}=\dfrac{1}{8}$

$\rm\longrightarrow\:\dfrac{1}{v}=\dfrac{1}{8}-\dfrac{1}{12}$

$\rm\longrightarrow\:\dfrac{1}{v}=\dfrac{3-2}{24}$

$\rm\longrightarrow\:\dfrac{1}{v}=\dfrac{1}{24}$

$\rm\longrightarrow\:v=24$

Hence,the image is formed at a distance of 24 cm from the convex lens.

Now,

we know that,

$\orange{\bigstar}\:{\underline{\boxed{\bf\blue{magnification\:(m)=\dfrac{v}{u}}}}}$

$\rm\longrightarrow\:m=\dfrac{24}{-12}$

$\rm\longrightarrow\:m=-2$

Hence,the image is real and inverted.

Then,

we know that,

$\pink{\bigstar}\:{\underline{\boxed{\bf\purple{m=\dfrac{h_2}{h_1}}}}}$

$\rm\longrightarrow\:-2=\dfrac{h_2}{7}$

$\rm\longrightarrow\:h_2=-2\times\:7$

$\rm\longrightarrow\:h_2=-14\:cm$

Thus,the height or size of the image is 14 cm and image is real and inverted.