Question

what is the focal length of double cancave lens kept in air with two sphercial surfaces of radii r1=30 and r2=60 cm take refractive index of lens n=15​

Answer 1

$\large\underline{\bigstar \: \: {\sf Correct \: Question - }}$

what is the focal length of double cancave lens kept in air with two sphercial surfaces of radii ${\sf R_1}$=30 and ${\sf R_2}$=60 cm take refractive index of lens μ= 1.5

$\large\underline{\bigstar \: \: {\sf Given-}}$

➩ In double concave lens

• Radii ${\sf R_1=-30\:cm\:R_2=+60cm}$
• Refractive index ( μ ) = 1.5

Sign Conversion

${\sf R_1}$ is left side thats why is negative (-)

${\sf R_2}$ is right side thats why is positive (+)

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Focal Lenght (f)

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\implies\underline{\boxed{\sf \dfrac{1}{f}=(\mu-1)\left[\dfrac{1}{R_1}-\dfrac{1}{R_2}\right]}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

$\implies{\sf (1.5-1)\left[\dfrac{1}{-30}-\left(\dfrac{1}{+60}\right)\right] }$

$\implies{\sf 0.5\left[\dfrac{1}{-30}-\dfrac{1}{60}\right]}$

$\implies{\sf 0.5\left[\dfrac{60-(-30)}{-30×60}\right]}$

$\implies{\sf 0.5\times\dfrac{90}{-1800}}$

$\implies{\sf 0.5\times (-0.05) }$

$\implies{\sf \dfrac{1}{f}=-0.025\:cm}$

$\implies{\sf f=\dfrac{1}{-0.025} }$

$\implies{\bf f=-40\:cm}$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Focal Lenght of double concave lens is ${\bf -40\:cm}$