A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror.The focal length of this mirror isa)- 30 cmb)- 20 cmc)- 40 cmd)- 60 cmCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Answer

H0=10mm hi=5mm v=−30cm

Here given ,

Now we know

m=−v u =hi ho ⇒u=−vho hi =−30*10 5 =−60 cm Now using mirror formula 1 f =1 v +1 u =1 −30 +1 −60 =−1 20 Hence f=−20 cm

Answer

2

Answer

Given data:
Length of the awl pin (object) = 10 mm
Length of the image = 5 mm
Distance of the image from the mirror = 30 cm

To find: Focal length of the mirror

Formula:
1/f = 1/v + 1/u
where,
f = focal length of the mirror
v = distance of the image from the mirror
u = distance of the object from the mirror

Calculation:
Length of the object = Length of the image
Therefore, the magnification produced by the mirror can be given as:
m = v/u = 5/10 = 1/2

Substituting the given values in the formula:
1/f = 1/v + 1/u
1/f = 1/30 + 1/(2u)
1/f = (2u + 30)/60u

We know that magnification (m) can also be given as:
m = -v/u

Substituting the given values in this formula:
1/2 = -v/10
v = -5 cm (negative sign indicates that the image is formed behind the mirror)

Substituting this value of v in the earlier formula:
1/f = 1/30 + 1/(2u)
1/f = (2u - 5)/60u

Substituting the value of magnification (m) in this formula:
1/f = (2/m - 5)/60(10/m)
1/f = (2m - 5)/600

Simplifying and solving for f:
f = 20 cm

Therefore, the focal length of the mirror is 20 cm. Answer: (b)

Answer

Answer

V=-30cm,ho=10mm=1cm,hi=5mm=0.5cm,f=?,we know that-->hi/ho=-v/u,0.5/1=-(-30)/u,u=60cm,by mirror formula-->1/f=1/v-1/u,1/f=1/30-1/(60),1/f=3/(-60),f=-60/3=-20 cm

Answer

-20 is correct answer

Answer

B) -20 cm is the answer

Answer

Height of the object = 10mm ->1cm
height if the image =( -5mm)->-0.5cm
m=hi/ho, v=(-30cm)
m=( -0.5)/1=(-0.5)
m=(-v/u)
(-0.5)=-(-30)/u
u=(-60cm)
1/v+1/u=1/f
1/-30+(-1/60)=1/f
f=(-20cm)

Answer

-20 cm or 200mm

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