In a concave mirror experiment, an object is placed at a distance ${x}_{1}$from the focus and the image is formed at a distance${x}_{2}$ from the focus. The focal length of the mirror would be
A.${x}_{1}{x}_{2}$
B.$\sqrt {{x}_{1}{x}_{2}}$
C.$\dfrac {{x}_{1}+{x}_{2}}{2}$
D.$\sqrt {\dfrac{{x}_{1}}{{x}_{2}}}$
Answer
634.5k+ views
Hint: Using the mirror formula this numerical can be solved. To solve this problem, you should know the difference between the mirror formula for concave mirror and convex mirror. Substitute the values for object distance and image distance. And thus, find the value for focal length of the concave mirror.
Formula used:
$\dfrac {1}{v}+ \dfrac {1}{u}= \dfrac {1}{f}$
Complete answer:
Given: Object distance= $({x}_{1}+f)$
Image distance = $({x}_{2}+f)$
Focal length = f
Using mirror formula,
Where, v is the distance of the image from the mirror
u is the distance of the object from the mirror
f is the focal length of the mirror
$\dfrac {1}{v}+ \dfrac {1}{u}= \dfrac {1}{f}$
Substituting values in above equation we get,
$\dfrac {1}{({x}_{2}+f)}+ \dfrac {1}{({x}_{1}+f)}= \dfrac {1}{f}$
$\Rightarrow \dfrac {1}{f}= \dfrac {({x}_{1}+f)+ ({x}_{2}+f)} {({x}_{1}+f)({x}_{2}+f)}$
$\Rightarrow f= \sqrt {{x}_{1}{x}_{2}}$
Thus, the focal length of the mirror would be $\sqrt{{x}_{1}{x}_{2}}$.
Hence, the correct answer is option B i.e. $\sqrt {{x}_{1}{x}_{2}}$
Note:
For solving such types of questions you should know the sign conventions. For a concave mirror, if the image and object are on the same side and in front of the mirror then the image distance is positive. But if the image is formed behind the mirror that is on the opposite side of the mirror and object then the image distance is negative. Focal length of the concave mirror is positive and focal length of the convex mirror is negative.
Formula used:
$\dfrac {1}{v}+ \dfrac {1}{u}= \dfrac {1}{f}$
Complete answer:
Given: Object distance= $({x}_{1}+f)$
Image distance = $({x}_{2}+f)$
Focal length = f
Using mirror formula,
Where, v is the distance of the image from the mirror
u is the distance of the object from the mirror
f is the focal length of the mirror
$\dfrac {1}{v}+ \dfrac {1}{u}= \dfrac {1}{f}$
Substituting values in above equation we get,
$\dfrac {1}{({x}_{2}+f)}+ \dfrac {1}{({x}_{1}+f)}= \dfrac {1}{f}$
$\Rightarrow \dfrac {1}{f}= \dfrac {({x}_{1}+f)+ ({x}_{2}+f)} {({x}_{1}+f)({x}_{2}+f)}$
$\Rightarrow f= \sqrt {{x}_{1}{x}_{2}}$
Thus, the focal length of the mirror would be $\sqrt{{x}_{1}{x}_{2}}$.
Hence, the correct answer is option B i.e. $\sqrt {{x}_{1}{x}_{2}}$
Note:
For solving such types of questions you should know the sign conventions. For a concave mirror, if the image and object are on the same side and in front of the mirror then the image distance is positive. But if the image is formed behind the mirror that is on the opposite side of the mirror and object then the image distance is negative. Focal length of the concave mirror is positive and focal length of the convex mirror is negative.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

