Respuesta :
For given diverging lens(Concave) ; a) The distance of the image is -6 Cm, b) The image is virtual.
What is diverging lens?
The lens that diverge incident parallel beam of light after refraction from the lens, is called diverging or concave lens.
- The images formed by it are always smaller, upright, towards object and between focus and optical center of the lens.
Given data:
a)
- u = -15 Cm, f = -10 Cm, v= ?
- by using lens formula,
[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\[/tex]
- On putting the values,
[tex]\frac{1}{-10} = \frac{1}{v} - \frac{1}{-15}\\\frac{1}{-10} = \frac{1}{v} + \frac{1}{15}[/tex]
[tex]\frac{1}{v} = \frac{1}{-10} -\frac{1}{15} \\\frac{1}{v} = \frac{-3-2}{30} \\[/tex]
[tex]v = -6 Cm[/tex]
b) Magnification ,
m= ([tex]\frac{v}{u}[/tex])
on putting values,
[tex]m =(\frac{-6}{-15} )=\frac{2}{5} =0.4[/tex]
since m is positive it means the image is virtual
Hence a) distance of image is -6 Cm and b) the image is virtual.
Find out more information about the concave lens here:
https://brainly.com/question/2919483
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