A biconvex lens of refractive index - 1 - is immersed in a liquid of refractive index - 2 - If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice is

The correct option is B μ _1 μ_1→μ_2 1f=( μ_eoq 1)( 1R_1 1R_2) divergent⇒μ_1μ_2 1 lt; 0⇒μ_1 lt;μ_2 since biconvex⇒( 1R_1 1R_2 ) gt; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Lens Maker's Formula

A biconvex le...

Question

A biconvex lens of refractive index $μ_{1}$ , is immersed in a liquid of refractive index $μ_{2}$ . If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice is

μ_{1} > μ_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

μ_{1} < μ_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

μ_{1} = μ_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

μ_{2} = \frac{μ_{1}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $μ_{1} < μ_{2}$
$μ_{1} \to μ_{2}$

$\frac{1}{f} = (μ_{e o q} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$d i v e r g e n t \Rightarrow \frac{μ_{1}}{μ_{2}} - 1 < 0 \Rightarrow μ_{1} < μ_{2}$

since $b i c o n v e x \Rightarrow (\frac{1}{R_{1}} - \frac{1}{R_{2}}) > 0$

Suggest Corrections

Similar questions

Q. A double convex lens, lens made of a material of refractive index

μ_{1}

is placed inside two liquids or refractive indices

μ_{2}

and

μ_{3}

, as shown

μ_{2} > μ_{1} > μ_{3}

. A wide, parallel beam of light is incident on the lens from the left.

An equi convex lens is made of a material of refractive index $μ_{1}$ and is placed inside two liquids of refractive indices $μ_{2}$ and $μ_{3}$ as shown with $μ_{2} > μ_{1} > μ_{3}$ . A wide, parallel beam of light is incident on the lens from the left. The lens will give rise to:

What is the relation between refractive indices $μ_{1}, μ_{2} and μ_{3}$ if the behaviour of light rays is as shown in the figure. The refractive index of medium to the left of the lens is $μ_{1}$ , that of lens is $μ_{3}$ and that of medium to the right of lens is $μ_{2}$ .

Q. A rectangular slab

A B C D

of refractive index

μ_{1}

is immersed in a liquid of refractive index

μ_{2}

(

μ_{2} < μ_{1})

. A ray of light is incident on the surface

A B

of the slab as shown in the figure.The maximum value of incidence such that the ray comes out only from the surface

C D

, is given by

Q. A right angled prism of refractive index

μ_{1}

is placed in a rectangular block of refractive index

μ_{2}

, which is surrounded by a medium of refractive index

μ_{3},

as shown in the figure. A ray of light

^{'} e^{'}

enters the rectangular block at normal incidence. Depending upon the relationship between

μ_{1}, μ_{2}

and

μ_{3},

it takes one of the four possible paths

^{'} e f^{'},^{'} e g^{'},^{'} e h^{'}

^{'} e i^{'}

Match the paths in

List -I

with conditions of refractive indices in

List -II

and choose the correct answer.
List - I List - II

$(P) e \to f$	$(1) μ_{1} > \sqrt{2} μ_{2}$
$(Q) e \to g$	$(2) μ_{2} > μ_{1} & μ_{2} > μ_{3}$
$(R) e \to h$	$(3) μ_{1} = μ_{2} = μ_{3}$
$(S) e \to i$	$(4) μ_{2} < μ_{1} < \sqrt{2} μ_{2} & μ_{2} > μ_{1}$

Join BYJU'S Learning Program

A biconvex lens of refractive index μ1 , is immersed in a liquid of refractive index μ2 . If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice is

The correct option is B μ1<μ2μ1→μ21f=(μeoq−1)(1R1−1R2)divergent⇒μ1μ2−1<0⇒μ1<μ2since biconvex⇒(1R1−1R2)>0

A biconvex lens of refractive index $μ_{1}$ , is immersed in a liquid of refractive index $μ_{2}$ . If a beam of parallel rays incident on the lens and the emergent rays are divergent then the correct choice is

The correct option is B $μ_{1} < μ_{2}$
$μ_{1} \to μ_{2}$

$\frac{1}{f} = (μ_{e o q} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$d i v e r g e n t \Rightarrow \frac{μ_{1}}{μ_{2}} - 1 < 0 \Rightarrow μ_{1} < μ_{2}$

since $b i c o n v e x \Rightarrow (\frac{1}{R_{1}} - \frac{1}{R_{2}}) > 0$