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CBSE Class 10 Science 2020 Delhi Set 1

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Question : 29 of 30
Marks: +1, -0
Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed :
(i) between optical centre and principal focus of a convex lens.
(ii) anywhere in front of a concave lens.
(iii) at 2F2 F of a convex lens.
State the signs and values of magnifications in the above mentioned cases (i) and (ii).
OR
An object 4.0 cm4.0\,\mathrm{cm} in size, is placed 25.0 cm25.0\,\mathrm{cm} infront of a concave mirror of focal length 15.0 cm15.0\,\mathrm{cm}.
(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(ii) Find the size of the image.
(iii) Draw a ray diagram to show the formation of image in this case.
Solution:  
Sign of magnification == Positive
Value of magnification == Greater than 1
Sign of magnification == Positive
Value of magnification == Greater than 1
OR
Given, size of object, Ho=4 cmH_o = 4\,\mathrm{cm}
Object distance, u=−25 cmu = -25\,\mathrm{cm}
Focal length of concave mirror, f=−15 cmf = -15\,\mathrm{cm}
(i) As per formula,
  1f  =  1v+  1u\;\frac{1}{f}\;=\;\frac{1}{v}+\;\frac{1}{u}
⇒      1−15  =  1v+  1−25\Rightarrow\;\; \;\frac{1}{-15}\;=\;\frac{1}{v}+\;\frac{1}{-25}
⇒      1v  =  1−15−  1−25=  1−15+  125\Rightarrow\;\; \;\frac{1}{v}\;=\;\frac{1}{-15}-\;\frac{1}{-25}=\;\frac{1}{-15}+\;\frac{1}{25}
  =  −25+15375=  −10375\;=\;\frac{-25+15}{375}=\;\frac{-10}{375}
  =  1−37.5\;=\;\frac{1}{-37.5}
⇒      1v  =  1−37.5\Rightarrow\;\; \;\frac{1}{v}\;=\;\frac{1}{-37.5}
⇒    v  =−37.5 cm\Rightarrow\;\; v\;=-37.5\,\mathrm{cm}
∴\therefore Image is formed on object side at 37.5 cm\mathrm{cm} from pole of concave mirror.
(ii) The formula for the size of an image
HiHO=−vu\frac{H_i}{H_O} = -\frac{v}{u}
Where HiH_i is the size of image.
Now,
  Hi4 cm  =  −(−37.5 cm)(−25 cm)\;\frac{H_i}{4\,\mathrm{cm}}\;=\;\frac{-(-37.5\,\mathrm{cm})}{(-25\,\mathrm{cm})}
Hi  =  −4×37.525 cm=−6 cmH_i\;=\;\frac{-4 \times 37.5}{25\,\mathrm{cm}} = -6\,\mathrm{cm}
∴\therefore Image is formed on object side at 37.5 cm\mathrm{cm} from pole of concave mirror.
(ii) The formula for the size of an image
HiHO=−vu\frac{H_i}{H_O} = -\frac{v}{u}
Where HiH_i is the size of image.
Now,
  Hi4 cm  =  −(−37.5 cm)(−25 cm)\;\frac{H_i}{4\,\mathrm{cm}}\;=\;\frac{-(-37.5\,\mathrm{cm})}{(-25\,\mathrm{cm})}
Hi  =  −4×37.525 cm=−6 cmH_i\;=\;\frac{-4 \times 37.5}{25\,\mathrm{cm}} = -6\,\mathrm{cm}
∴\therefore Size of image =6 cm=6\,\mathrm{cm}
Hence, screen should place 37.5 cm37.5\,\mathrm{cm} left side of mirror and size of image is 6 cm6\,\mathrm{cm}. It is inverted and larger than object's size.
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