Question

A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of
focal length 20 cm. The distance of the object from the lens is 30 cm. Find the
(i) positive (ii) nature and (iii) size of the image formed.
oth
A
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Answer 1

Given :-

• Object height (ho) = + 5 cm
• Focal length (f) = + 20cm
• Object distance (u) = - 30cm

To find :-

• Position
• Nature
• Size of image

Solution :-

Apply lens formula

→ 1/v - 1/u = 1/f

→ 1/v - (-1)/30 = 1/20

→ 1/v + 1/30 = 1/20

→ 1/v = 1/20 - 1/30

→ 1/v = 3 - 2/60

→ 1/v = 1/60

→ v = 60 cm

• Hence, image distance is 60 cm

Now, according to magnification of lens

→ m = v/u = hi/ho

→ v/u = hi/ho

→ -60/30 = hi/5

→ - 2 = hi/5

→ hi = - 10cm

Hence,

• Height of image = - 10cm
• Nature of image = Real and inverted

Answer 2

Solution :

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

• Focal length of lens, (f) = 20 cm
• Distance of object from lens, (u) = -30 cm
• Height of object, (h1) = 5 cm

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

As we know that formula of the lens;

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

$\mapsto\sf{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }\\\\\\\mapsto\sf{\dfrac{1}{20} =\dfrac{1}{v} -\dfrac{1}{(-30)} }\\\\\\\mapsto\sf{\dfrac{1}{20} =\dfrac{1}{v} +\dfrac{1}{30} }\\\\\\\mapsto\sf{\dfrac{1}{v} =\dfrac{1}{20} -\dfrac{1}{30} }\\\\\\\mapsto\sf{\dfrac{1}{v} =\dfrac{3-2}{60} }\\\\\\\mapsto\sf{\dfrac{1}{v} =\dfrac{1}{60} }\\\\\\\mapsto\bf{v=60\:cm}$

∴ Image is 60 cm far from the lens on other side,i.e, behind the lens image is real & Inverted.

Now;

As we know that formula of the linear magnification;

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

$\longrightarrow\sf{m=\dfrac{h_2}{h_1} =\dfrac{v}{u} }\\\\\\\longrightarrow\sf{\dfrac{h_2}{5} =\cancel{\dfrac{60}{-30} }}\\\\\\\longrightarrow\sf{\dfrac{h_2}{5} =-2}\\\\\longrightarrow\sf{h_2 = -2\times 5}\\\\\longrightarrow\bf{h_2 = -10\:cm}$

Thus;

Image is inverted & a height of image will be 10 cm .