Question

An obiect of height 2 cms kept at a
distance of 30 cm from a convex lens
of focal length 20 cm Find the
position & nature of the image​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Height of Object from lens, (h1) = 2 cm
• Distance of object from lens, (u) = -30 cm
• Focal length of the lens, (f) = 20 cm

$\underline{\bf{Explanation\::}}$

As we know that formula of the lens;

$\boxed{\bf{\dfrac{1}{f} = \dfrac{1}{v} -\dfrac{1}{u} }}$

A/q

$\mapsto\tt{\dfrac{1}{f} =\dfrac{1}{v} - \dfrac{1}{u} }$

$\mapsto\tt{\dfrac{1}{20} =\dfrac{1}{v} - \dfrac{1}{(-30)} }$

$\mapsto\tt{\dfrac{1}{20} =\dfrac{1}{v} + \dfrac{1}{30} }$

$\mapsto\tt{\dfrac{1}{v} =\dfrac{1}{20} - \dfrac{1}{30} }$

$\mapsto\tt{\dfrac{1}{v} =\dfrac{3 - 2}{60}}$

$\mapsto\tt{\dfrac{1}{v} =\dfrac{1}{60}}$

$\mapsto\bf{v = 60\:cm}$

∴ The distance formed of Image from lens will be 60 cm .

As we know that formula of the magnification;

$\mapsto\bf{m = \dfrac{Height \:of\:Image\:(I)}{Height\:of\:Object\:(O)} = \dfrac{Distance\:of\:image}{Distance\:of\:Object} = \dfrac{v}{u} }$

Now,

$\longrightarrow\tt{m = \dfrac{h_2}{h_1} = \dfrac{-v}{u} }$

$\longrightarrow\tt{ \dfrac{h_2}{2} = \dfrac{-60}{-30} }$

$\longrightarrow\tt{ \dfrac{h_2}{2} = \cancel{\dfrac{-60}{-30}} }$

$\longrightarrow\tt{ \dfrac{h_2}{2} = 2}$

$\longrightarrow\tt{h_2 = 2 \times 2}$

$\longrightarrow\bf{h_2 = 4\:cm}$

Thus,

The Image is virtual & erect and size of Image will be 4 cm .

Answer 2

${\tt{\purple{\underline{\underline{\huge{Answer:}}}}}}$

here is your answer

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${\tt{\pink{\underline{\underline{\huge{Hope-it-helps}}}}}}$