Question

question in the attachment.
thanks for helping!!​

Answer 1

Given:-

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

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 .

Answer 2

Answer:

Given:-

Focal length of lens, (f) = 20 cm

Distance of object from lens, (u) = -30 cm

Height of object, (h1) = 5 cm

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 .