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If a concave mirror has a focal length of 10cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.

Posted by Shrey Raj (Jan 12, 2017 1:10 p.m.) (Question ID: 1308)

• Given f = -10 cm

magnification is twice so taking

m = +2 and -2m = -v/u

2 = -v/u

v = -2u

1/f = 1/v + 1/u

-1/10 = -1+2/2u

u = -5

Now taking magnification as -2

m = -v/u

-2 = -v/u

v = 2u

1/f = 1/v + 1/u

-1/10 = (1+2)/2u

-2u = 30 or

u = -15

Answered by Naveen Sharma (Jan 12, 2017 2:02 p.m.)
Thanks (1)
• Ans.

The image twice the height of the object can

A) Virtual,erect and enlarged

Or

B) Real,inverted and enlarged

1) Virtual,erect and enlarge,

in this we will take m = 2

now, m = - v / u
=>2 = - v / u
=> 2u = -v

therefore, v = -2u

1/f=(1/v)+(1/u )
=> 1/(-10)=1/(-2u) + 1/u
=> 1/-10=(-1+ 2)/2u
=> 1/-10=1/2u
-10= 2u, u=-5 cm

B) Real,inverted and enlarge ,

in this we will take,  m = -2

now,m = - v / u
-2 = - v / u
- 2u = -v

therefore , v = 2u
1 / f  = (1 / v )+ (1 / u)
1 / (-10)  = 1 / (2u) + 1 / u
1 / - 10    = (1 + 2) / 2u
1 / - 10  =  3 / 2u
- 30 = 2u therefore , u  = - 15 cm

Therefore, an image twice the height of the object can be obtained when the object is placed at

A) 5cm from the concave mirror

B) 15cm from the concave mirror

Answered by Payal Singh (Jan 12, 2017 2 p.m.)
Thanks (1)