Since, the given lens is a convex lens hence it should converge a parallel beam of light falling on it to a single focal point. However, since it is showing the exactly opposite behaviour, thus it is definitely placed in a medium of higher refractive index.Thus, we conclude that \$ {\mu}_{1} > {\mu} \$

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If \$\mu \$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

\$\textbf {Hint :}\$ Use critical angle \$C=\sin^{-1}\left(\dfrac{1}{\mu}\right)\$\$\textbf{Step 1: }\$ Calculating the maximum value of the base angle \$\alpha\$If \$\alpha\$ is maximum value of base angle for which light is totally reflected from the hypotenuse and \$\theta\$ is the minimum value of angle of incidence at hypotenuse for total internal reflection.Then ,\$ \theta =sin^{-1}\left ( \dfrac{1}{\mu}\right )\$From diagram\$\alpha +\theta =90^o\$\$\Rightarrow \alpha = 90^o-sin^{-1}\left ( 1/\mu \right )=cos^{-1}\left ( 1/\mu \right )\$\${\textbf{Correct option: D}}\$

A double convex lens, made of a material of refractive index \$\mu _{1}\$ , is placed inside two liquids of refractive indices \$\mu _{2}\$ and \$\mu _{3}\$ , where \$\mu _{2}>\mu _{1}>\mu _{3}\$. A wide , parallel beam of light is incident on the lens from the left .The lens will give rise to

\$\therefore\$ analyzing the rays it gives a divergent ray above the axis andconvergent ray below \$\mu_2 > \mu_1 > \mu_3\$from medium 2\$\Rightarrow\$ when light ray enters into medium \$1 (\mu_1)\$ then moves away from normalfrom medium 3\$\Rightarrow\$ when light ray enters into medium \$1 (\mu_1)\$ the moves towards the normal.

The refractive index of the material of a concave lens is \$\mu \$ and is placed in the medium of refractive index \$\mu ^{1}\$. If \$\mu ^{1}>\mu \$, then which of the following ray diagram is correct:

Since, the tendency of a concave lens is to diverge a parallel beam of light falling on the lens surface, it would show exactly the opposite behaviour if placed in a medium of higher refractive index than the material of the lens.Thus, the correct option will be one in which a concave lens diverges a parallel beam of a light ray.

Solve Study Textbooks Guides

Use app

Ray Optics And Optical Instruments

Q: A light ray is incident perpendicularly to one face of a 90\$^{\circ}\$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45\$^{\circ}\$, we conclude that the refractive index \$n\$ :

We know\$\theta _{cric}=sin^{-1}(\dfrac{1}{n })\$\$45^{o}> \theta \$\$sin(45^o)> \dfrac{1}{n}\$ \$\Rightarrow n > \sqrt{2} \$

Q: A ray of light falls on the surface of a spherical glass paperweight making an angle \$a\$ with the normal and is refracted in the medium at angle \$b\$. The angle of deviation of the emergent ray from the direction of incident ray is :

From the above figure we can see that \$\theta =2(a-b)\$Exterior angle=sum of interior opposite angles. So answer is option B.

Q: Match the elements of table I and II.Table ITable IIa) Myopiai) Bifocal lenb) Hypermetropiaii) Cylindrical lensc) Presbyopiaiii) Concave lensd) Astigmationiv) Convex lens

Myopia is inability to see distant objects as parallel rays from distant objects converge to a point in front of retina due to increased power of eye lens.Hence it is corrected by a concave lens of negative power to decrease net power.Hyper metropia is inability to see near objects as diverging rays from near objects can not be converged to retina due to decreased power of eye lens.Hence a convex lens of positive power is used to increase power.Presbyopia is caused by changes in the lens inside the eye. As people age, the lens becomes harder and less elastic, making it more difficult for the eye to focus on close objects.It is corrected by bifocal lenses.Astigmatism is a type of refractive error in which the eye does not focus light evenly on the retina. It can be corrected by cylindrical lenses.

Q: Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

\$\begin{array}{l}\text { Ths is a combination of convex, concave and convex } \\\text { lens. Number them } 1,2,3 \\\qquad \begin{aligned}\text { given } f_{1} &=f_{3}=20 \mathrm{~cm} \\\mu_{1} &=\mu_{3}=1.5 \\\mu_{2} &=4/3\end{aligned}\end{array}\$\$\begin{array}{l}\quad \mu_{2}=4 / 3 \\\text { Now densmaker formula } \frac{1}{f}=\left(\mu_{r}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\text { for (1) } \quad \frac{1}{f_{1}}=(1.5-1)\left(\frac{1}{R_{1}}-\left(-\frac{1}{R}\right)\right)\end{array}\$\$\text { (equi convex lenS }\left.R_{1}=R_{2}\right)\$\$\begin{array}{l}\frac{1}{20}=\frac{1}{2}\times \frac{2}{R} \\ R=20 \mathrm{~cm} \\\text { for (2) } \\\qquad \begin{array}{l}\frac{1}{f_{2}}=\left(\frac{4}{3}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\frac{1}{f_{2}}=\frac{1}{3}\left(\frac{-2}{R}\right) \\ f_{2}=-30\mathrm{~cm}\end{array}\end{array}\$\$\begin{array}{l}\text { Now we know that } \\\qquad \begin{aligned}\text { ieffective } &=P_{1}+P_{2}+P_{3} \\\frac{1}{f_{e q}} &=\frac{1}{20}-\frac{1}{30}+\frac{1}{20} \\ f_{\text {eq }}=15\mathrm{~cm}\end{aligned} \\\text { So it is a converging lens of focal length } 15 \mathrm{~cm} .\end{array} \$

Q: A ray incident at a point at an angle of incidence \$60\$\$^{\circ}\$ enters a glass sphere of \$\mu =\sqrt{3}\$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

Using Snell's Law at point \$P\$ we get:-\$\dfrac{sin\ 60^o}{sin\ r_1}=\sqrt3\$\$sin \ r_1= \dfrac12\implies r_1= 30^o\$As we know that \$r_1=r_2 \implies r_2=30^o\$Using Snell's Law at point \$Q\$ we get:-\$\dfrac{sin \ 30^o}{sin\ i_2}=\dfrac1{\sqrt3}\implies i_2= 60^o\$Since reflection at point \$Q\$ occurs\$\implies r_2'=r_2= 60^o\$Let angle between the refracted ray and reflected ray be \$\alpha\$\$\alpha= 180^o- (r_2'+r_2)= 90^o\$Hence option \$(B)\$ is correct.

Q: A defective eye cannot see close objects clearly because their image is formed:

A defective eye cannot see close objects clearly because its image is formed beyond the retina. This defect is known as farsightedness. Here rays from near objects could not be converted at the retina due to focal length of the eye lens could not be decreased, so, image is formed beyond the retina.

Physics
1901 Views

Easy Qs Med Qs Hard Qs

A light ray is incident perpendicularly to one face of a 90 $^{\circ}$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45 $^{\circ}$ , we conclude that the refractive index $n$ :

n > \frac{1}{2}

n > 2

n < \frac{1}{2}

n < 2

Medium

View solution>

In the diagram shown

μ^{1} = μ

μ^{1} < μ

μ^{1} > μ

μ^{1} \geq μ

Medium

View solution>

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $μ$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

s i n^{- 1} (\frac{1}{μ})

t a n^{- 1} (\frac{1}{μ})

s i n^{- 1} (\frac{μ - 1}{μ})

c o s^{- 1} (\frac{1}{μ})

Medium

View solution>

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

a single convergent beam

two different convergent beams

two different divergent beams

a convergent and a divergent beam

Medium

View solution>

A ray of light falls on the surface of a spherical glass paperweight making an angle $a$ with the normal and is refracted in the medium at angle $b$ . The angle of deviation of the emergent ray from the direction of incident ray is :

a - b

2 (a - b)

\frac{( a - b )}{2}

b - a

Medium

View solution>

Match the elements of table I and II.

Table I Table II
a) Myopia i) Bifocal len
b) Hypermetropia ii) Cylindrical lens
c) Presbyopia iii) Concave lens
d) Astigmation iv) Convex lens

a - iii, b - iv, c- i, d -ii

a - iv, b - iii, c -i, d - ii

a - i, b - ii, c - iii, d - iv

a - ii, b - iv, c -i, d - iii

Medium

View solution>

Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

converging lens of focal length 50 cm.

diverging lens of focal length 15 cm

converging lens of focal length 15 cm

diverging lens of focal length 50 cm

Medium

View solution>

The refractive index of the material of a concave lens is $μ$ and is placed in the medium of refractive index $μ^{1}$ . If $μ^{1} > μ$ , then which of the following ray diagram is correct:

None of these

Medium

View solution>

A ray incident at a point at an angle of incidence $60$ $^{\circ}$ enters a glass sphere of $μ = 3$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

50

^{\circ}

90

^{\circ}

60

^{\circ}

40

^{\circ}

Medium

View solution>

A defective eye cannot see close objects clearly because their image is formed:

On the eye lens

Between eye lens and retina

On the retina

Beyond retina

Medium

View solution>

Table I	Table II
a) Myopia	i) Bifocal len
b) Hypermetropia	ii) Cylindrical lens
c) Presbyopia	iii) Concave lens
d) Astigmation	iv) Convex lens

Ray Optics And Optical Instruments

Physics 1901 Views

A light ray is incident perpendicularly to one face of a 90∘ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45∘, we conclude that the refractive index n :

In the diagram shown

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If μ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

A double convex lens, made of a material of refractive index μ1​ , is placed inside two liquids of refractive indices μ2​ and μ3​ , where μ2​>μ1​>μ3​. A wide , parallel beam of light is incident on the lens from the left .The lens will give rise to

A ray of light falls on the surface of a spherical glass paperweight making an angle a with the normal and is refracted in the medium at angle b. The angle of deviation of the emergent ray from the direction of incident ray is :

Match the elements of table I and II. Table ITable IIa) Myopiai) Bifocal lenb) Hypermetropiaii) Cylindrical lensc) Presbyopiaiii) Concave lensd) Astigmationiv) Convex lens

Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

The refractive index of the material of a concave lens is μ and is placed in the medium of refractive index μ1. If μ1>μ, then which of the following ray diagram is correct:

A ray incident at a point at an angle of incidence 60∘ enters a glass sphere of μ=3​ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

A defective eye cannot see close objects clearly because their image is formed:

Physics
1901 Views

A light ray is incident perpendicularly to one face of a 90 $^{\circ}$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45 $^{\circ}$ , we conclude that the refractive index $n$ :

A ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $μ$ is the refractive index of the material of the prism, the maximum value of the base angle for which light is totally reflected from the hypotenuse is :

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

A ray of light falls on the surface of a spherical glass paperweight making an angle $a$ with the normal and is refracted in the medium at angle $b$ . The angle of deviation of the emergent ray from the direction of incident ray is :

Match the elements of table I and II.

Table I Table II
a) Myopia i) Bifocal len
b) Hypermetropia ii) Cylindrical lens
c) Presbyopia iii) Concave lens
d) Astigmation iv) Convex lens

The refractive index of the material of a concave lens is $μ$ and is placed in the medium of refractive index $μ^{1}$ . If $μ^{1} > μ$ , then which of the following ray diagram is correct:

A ray incident at a point at an angle of incidence $60$ $^{\circ}$ enters a glass sphere of $μ = 3$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :