a. An object 1 cm high is placed at 10cm in front of a concave mirror of focal length 15 cm. Find the position, height, and nature of the image.
b. A point source S is placed midway between two converging mirrors having equal focal length f as shown in Figure. Find the value of d for which only one image is formed.

Explanation

a. Mirror formul: `(1)/(v)+(1)/(u)=(1)/(f)`
Here, `u=-10cm, f=-15cm`
`:. (1)/(v)=(1)/(10)-(1)/(15)=(15-10)/((15)(10))=(1)/(30)`
The positive sign indicates that the image is virtual and erect and it is formed behind the mirror.
Alternatively, we can apply Newton's formul.
Here, `x_(0)=u-f=-10-(-15)=5cm`
`x_(1) =(f^(2))/(x_(0))=((15)^(2))/(5)=45cm`
Since `x_(i)=v-f`, therefore
`v=x_(i)+f=45-15=30cm`
The magnification,
`m=-(v)/(u)=(h_(i))/(h_(0)) `
`h_(i)=h_(0)(-(v)/(u))=1(-(30)/(-10)) =3cm`