$A$ lens of refractive index '$\mu$' has focal length '$f$'. When the lens is immersed in a liquid of refractive index '$\mu_0$',its focal length becomes '$f_0$'. Then '$f_0$' is given by

  • A
    $\frac{(\mu_0 - \mu)f}{\mu(\mu_0 - 1)}$
  • B
    $\frac{\mu(\mu_0 - 1)f}{(\mu_0 - \mu)}$
  • C
    $\frac{(\mu - \mu_0)f}{\mu_0(\mu - 1)}$
  • D
    $\frac{\mu_0(\mu - 1)f}{(\mu - \mu_0)}$

Explore More

Similar Questions

$A$ concave lens of glass,refractive index $1.5$,has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

Consider the following statements regarding the real images formed with a converging lens.
$I$. Real images can be seen only if the image is projected onto the screen.
$II$. The real image can be seen only from the same side of the lens as that on which the object is positioned.
$III$. Real images produced by converging lenses are not only laterally but also longitudinally inverted as with mirrors.
Which of the above statement$(s)$ is/are incorrect?

What is a lens?

The distance between an object and its image produced by a converging lens is $0.72 \ m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04 \ m$ towards the lens?

The following figure shows a beam of light converging at point $P$. When a concave lens of focal length $16 \,cm$ is introduced in the path of the beam at a place shown by the dotted line such that $OP$ becomes the axis of the lens,the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to (in $\,cm$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo