Question

1. A 1.0-cm-tall object is 85cm in front of a converging lens that has a 35cm focal length. Calculate the image position and image height.

2. A 4.0-cm-tall object is 16cm in front of a diverging lens that has a -20cm focal length. Calculate the image position and image height.

3. A 4.0-cm-tall object is 14cm in front of a concave mirror that has a 25 cm focal length. Calculate the image position and image height.

Answer 1

1). Height of object, h_o= 1.0cm

Object Distance, u= 85cm

Focal length of converging lens, f= 35cm

Let image position be "v" and image height be "h_i", then using Lens Equation,

$\frac{1}{f}= \frac{1}{v}+\frac{1}{u}$

$\frac{1}{v}= \frac{1}{f}- \frac{1}{u}$

   $v= \frac{fu}{u-f}$

using given values in above,

$v= \frac{35 \times 85}{85-35}= 59.5cm$ (ANS)

Using the formula of magnification,

   $\frac{h_i}{h_o}= \frac{-v}{u}$

$\frac{h_i}{1.0}= \frac{-59.5}{85}$

   $h_i=-0.7cm$ (ANS)(Negative sign shows that image is real and in downwards direction)

2). Height of object, h_o= 4.0cm

Object Distance, u= 16cm

Focal length of diverging lens, f= -20cm

Let image position be "v" and image height be "h_i", then using Lens Equation,

$\frac{1}{f}= \frac{1}{v}+\frac{1}{u}$

$\frac{1}{v}= \frac{1}{f}- \frac{1}{u}$

   $v= \frac{fu}{u-f}$

using given values in above,

$v= \frac{16 \times (-20)}{16-(-20)}= -8.89cm$ (ANS)

Using the formula of magnification,

   $\frac{h_i}{h_o}= \frac{-v}{u}$

$\frac{h_i}{4.0}= \frac{-(-8.89)}{16}$

   $h_i=2.22cm$ (ANS) (Positive sign shows that the image is errect)

3). Height of object, h_o= 4.0cm

Object Distance, u= 14cm

Focal length of concave mirror, f= 25cm

Let image position be "v" and image height be "h_i", then using Lens Equation,

$\frac{1}{f}= \frac{1}{v}+\frac{1}{u}$

$\frac{1}{v}= \frac{1}{f}- \frac{1}{u}$

   $v= \frac{fu}{u-f}$

using given values in above,

$v= \frac{14 \times 25}{14-25}= -31.81cm$ (ANS)

Using the formula of magnification,

   $\frac{h_i}{h_o}= \frac{-v}{u}$

$\frac{h_i}{4.0}= \frac{-31.81}{14}$

   $h_i=-9.08cm$ (ANS) (Negative sign shows that image is real and in downwards direction)