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CBSE Class 10 Science 2017 Delhi Term II Set 1 Paper

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Question : 16 of 36
Marks: +1, -0
An object 4 cm4\text{ cm} in height, is placed at 15 cm15\text{ cm} in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.
Solution:  
Given : u=−15 cm,f=−10 cm,ho=4 cmu=-15\text{ cm}, f=-10\text{ cm}, h_o=4\text{ cm}
Using the mirror formula
1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
1v−115=−110\frac{1}{v}-\frac{1}{15}=\frac{-1}{10}
1v=−110+115\frac{1}{v}=\frac{-1}{10}+\frac{1}{15}
1v=−15+10150=−5150\frac{1}{v}=\frac{-15+10}{150}=\frac{-5}{150}
v=−30 cmv=-30\text{ cm}
Thus, to obtain a sharp image of the object, the screen should be placed in front of the mirror at a distance of 30 cm30\text{ cm}.
Now,
m=−vu=height of imageheight of objectm=\frac{-v}{u}=\frac{\text{height of image}}{\text{height of object}}
m=−(−30−15)⇒m=−2m=-\left(\frac{-30}{-15}\right)\Rightarrow m=-2
−2=hi4⇒hi=−8 cm-2=\frac{h_i}{4}\Rightarrow h_i=-8\text{ cm}
∴\therefore Height of the image is 8 cm8\text{ cm}. (inverted)
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