datavorous/entrance-exam-dataset huggingface dataset:
A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at pressure p and temperature T, whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be

Options:
A p2, T
B p2,Â T2
C p,Â T
D p,T2

The correct answer is:
p2, T

Solution:
Internal energy of the gas remains constant, hence
T2=T
Using p1V1=p2V2
p.V2=p2V
p2=p2

Topic: Physics, Thermodynamics, NEET
A continuous function fx is such that f3x=2fx, âxâR. If â«01fxdx=1, then â«13fxdx is equal to

Options:

The correct answer is:
5

Solution:
Given, f3x=2fx
Also, ∫01fxdx=1
⇒12∫01f3xdx=1
Substituting, 3x=t, dx=dt3, we get,
16∫03ftdt=1
⇒∫03ftdt=6
⇒∫01ftdt+∫13ftdt=6
⇒∫13ftdt=6-1=5

Topic: Mathematics, Definite Integration, JEE Main
A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of $10 \mathrm{~cm}$. The separation between the two lenses is $2 \mathrm{~cm}$. The focal lengths of the component lenses

Options:
A
$18 \mathrm{~cm}, 20 \mathrm{~cm}$

B
$10 \mathrm{~cm}, 12 \mathrm{~cm}$

C
$12 \mathrm{~cm}, 14 \mathrm{~cm}$

D
$16 \mathrm{~cm}, 18 \mathrm{~cm}$

The correct answer is:
$18 \mathrm{~cm}, 20 \mathrm{~cm}$

Solution:
For minimum spherical aberration separation, $d=f_1-f_2=2 \mathrm{~cm}$
Resultant focal length $=F=10 \mathrm{~cm}$
Using $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}$ and solving, we get $\mathrm{f}_1, \mathrm{f}_2 18 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A converging lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axes coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens, as shown in figure. It is found that the final beam comes out parallel to the principal axis. Let the separation between the mirror and the lens be 1 0 Ã k . Find k.

Options:

The correct answer is:
4

Solution:
Let us first locate the image of S formed by the lens L. Here u = 12 cm and f = 15 cm. We have,
1 vâ¡ - 1 uâ¡ = 1 f
or, 1 vâ¡ = 1 fâ¡ + 1 uâ¡
= 1 1 5 cm - 1 1 2 cm
or, vâ¡ = - 6 0 cm.
The negative sign shows that the image is formed to the left of the lens as suggested in the figure. The image I1 acts as the source for the mirror. The mirror forms an image I2 of the source I1. This image I2 then acts as the source for the lens and the final beam comes out parallel to the principal axis. Clearly I2 must be at the focus of the lens. We have,
I1 I2 = I1L + LI2 = 60 cm + 15 cm = 75 cm.
Suppose the distance of the mirror from I2 is x cm. For the reflection from the mirror,
u = MI1 = - (75 + x)cm, v = - x cm and f = - 20 cm.
Using 1 vâ¡ - 1 uâ¡ = 1 f ,
1 x + 1 7 5 + x = 1 2 0
or, 7 5 + 2 x 7 5 + x x = 1 2 0
or, x 2 + 3 5 x - 1 5 0 0 = 0
or, x = - 3 5 Â± 3 5 Ã 3 5 + 4 Ã 1 5 0 0 2 .
This gives x = 25 or - 60.
As the negative sign has no physical meaning, only positive sign should be taken. Taking x = 25, the separation between the lens and the mirror is (15 + 25) cm = 40 cm.
Hence k = 4

Topic: Physics, Ray Optics, JEE Main
A converging mirror is placed on the right hand side of a converging lens as shown in the figure. The focal length of the mirror and the lens are 20 cm and 15 cm respectively. The separation between the lens and the mirror is 40 cm and their principal axes coincide. A point source is placed on the principal axis at a distance d to the left of the lens. If the final beam comes out parallel to the principal axis, then the value of d is:

Options:
A 4 cm
B 8 cm
C 12 cm
D 16 cm

The correct answer is:
12 cm

Solution:
Image will act as object for mirror,

v1+u2=40

1f2=1u2+1v2

here v2=∞ ⇒f2=u2=15 cm

Lens formula,

1f1=1u1-1v1

∴ 120=1d-125 ⇒d≈12 cm

Topic: Physics, Ray Optics, JEE Main
A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point $15 \mathrm{~cm}$ from the lens on the opposite side. If the lens is removed the point where the rays meets will move $5 \mathrm{~cm}$ closer to the lens. The focal length of the lens is

Options:
A $-10 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $-30 \mathrm{~cm}$
D $5 \mathrm{~cm}$

The correct answer is:
$-30 \mathrm{~cm}$

Solution:
Given $u=10 \mathrm{~cm}, y=15 \mathrm{~cm}$
$$
\begin{aligned}
& \frac{1}{f}=\frac{1}{y}-\frac{1}{u} \\
& \frac{1}{f}=\frac{1}{15}-\frac{1}{10} \\
& \frac{1}{f}=\frac{10-15}{150} \\
& \frac{1}{f}=\frac{-5}{150} \\
& f=-30 \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2011 (Mains)
A convex and a concave lens separated by distance are then put in contact. The focal length of the combination

Options:
A decreases
B increases
C becomes 0
D remains the same

The correct answer is:
decreases

Solution:
When two lenses are placed coaxially at a distance from each other, then equivalent focal length ( ) is given by
$$
\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}
$$
When the lenses are put together.
$$
\mathrm{d}=0
$$
Hence, the focal length of the combination will decrease.

Topic: Physics, Ray Optics, KCET, KCET 2008
A convex and a concave lens separated by distance $d$ are then put in contact. The focal length of the combination

Options:
A decreases
B increases
C becomes zero
D remains the same

The correct answer is:
decreases

Solution:
Let $f_{1}$ and $f_{2}$ are focal lengths of convex and concave lens respectively, when they are kept in contact co-axially, then focal length $F$ of the combination is given as
$$
\begin{aligned}
\frac{1}{F} &=\frac{1}{f_{1}}+\frac{1}{\left(-f_{2}\right)} \quad\left[\begin{array}{l}
\text { Focal length of concave } \\
\text { lens is negative. }
\end{array}\right] \\
\Rightarrow \quad \frac{1}{F} &=\frac{f_{2}-f_{1}}{f_{1} f_{2}} \\
\Rightarrow \quad F &=\frac{f_{1} f_{2}}{f_{2}-f_{1}}
\end{aligned}
$$
When both lenses are kept at a distance $d$, then their combined focal length $F^{\prime}$ is given as
$$
\begin{aligned}
\frac{1}{F^{\prime}} &=\frac{1}{f_{1}}+\frac{1}{\left(-f_{2}\right)}-\frac{d}{f_{1}\left(-f_{2}\right)} \\
\Rightarrow \quad \frac{1}{F^{\prime}} &=\frac{1}{f_{1}}-\frac{1}{f_{2}}+\frac{d}{f_{1} f_{2}} \\
\Rightarrow \quad \frac{1}{F^{\prime}} &=\frac{f_{2}-f_{1}+d}{f_{1} f_{2}} \\
\Rightarrow \quad F^{\prime} &=\frac{f_{1} f_{2}}{f_{2}-f_{1}+d}
\end{aligned}
$$
From Eqs. (i) and (ii), it is clear that $F^{\prime} < F$ i.e. focal length of combination will decrease.

Topic: Physics, Ray Optics, COMEDK, COMEDK 2018
A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then the distance d in cm will be :

Options:
A 50
B 15
C 25
D 30

The correct answer is:
15

Solution:
1F=1f1+1f2-df1f2
∴1∞=120+1-5-d20-5
⇒0=120-15+d100
∴d100=15-120=20-5100
∴ d=15 cm

From figure it is clear that separation between lenses, d=20−5=15 cm

Topic: Physics, Ray Optics, JEE Main
A convex lens and a concave lens, each having same focal length of \( 25 \mathrm{~cm} \), are put in contact to form a combination of lenses. The power in dioptres of the combination is

Options:
A Zero
B \( 25 \)
C \( 50 \)
D Infinite

The correct answer is:
Zero

Solution:
The focal length of convex lens, f1=+25 cm and the focal length of the concave lens, f2=-25 cm.

Power of combination in dioptres of the combination of lenses is given by,

P=P1+P2=100f1+100f2=10025-10025=0.

Topic: Physics, Ray Optics, JEE Main
A convex lens forms inverted image of a real object on a fixed screen. The size of image is 9 cm. When lens is displaced 40 cm along principle axis it again forms a real image of size 4 cm on the screen. Focal length of the lens is .

Options:
A 48 cm
B 100 cm
C 30 cm
D 10 cm

The correct answer is:
48 cm

Solution:
Since m1 m2=1 ⇒ h1h0 h2h0=1
h0= h1 h2=6cm
for first position of the lens
m=-32=- x + 40x
⇒ 3x=2x+80
⇒ x= 80
u=-x, y=(40+x)

1f=1120-1-80

⇒f=48 cm

Topic: Physics, Experimental Physics, JEE Main
A convex lens has a focal length $f$. It is cut into two parts along the dotted line as shown in the figure. The focal length of each part will be

Options:
A $\frac{f}{2}$
B $f$
C $\frac{3}{2} f$
D $2 f$

The correct answer is:
$2 f$

Solution:
$f=\frac{R}{2(\mu-1)}, f^{\prime}=\frac{R}{(\mu-1)} \Rightarrow f^{\prime}=2 f$

Topic: Physics, Ray Optics, JEE Main
A convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ( $\mu=$ the refractive index of the material of the lens)

Options:
A $f$
B $\frac{f}{2}$
C $2 f$
D $(\mu-1) f$

The correct answer is:
$2 f$

Solution:
The given, $R_1=R, R_2=-R$
$$
f=F
$$
Lens Maker's formula
$$
\begin{aligned}
\frac{1}{F} & =(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\
\frac{1}{f} & =(\mu-1)\left[\frac{1}{R}+\frac{1}{R}\right] \\
f & =\frac{R}{2(\mu-1)} \\
R & =2 f(\mu-1)
\end{aligned}
$$
Now, it is divided vertically into two identical plano convex lens
$$
\begin{aligned}
& \frac{1}{f^{\prime}}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\
& \frac{1}{f_1}=(\mu-1)\left[\frac{1}{R}-\frac{1}{\infty}\right] \quad\left[\because R_1=R, R_2=\infty\right] \\
& f_1=\frac{R}{(\mu-1)} \\
& f_1=\frac{2 f(\mu-1)}{(\mu-1)} \\
& f_1=2 f
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
A convex lens is dipped in a liquid whose refractive index is equal to refractive index of lens material. Then its focal length will

Options:
A become infinite
B remain same
C decrease
D become zero

The correct answer is:
become infinite

Solution:
Given that refractive of liquid is equal to refractive index of lens.
$\Rightarrow n_{\text {lens }}=n_{\text {liquid }}$
Let the focal length be $f$.
From lens makers formula, we get $\frac{1}{f}=\frac{\left(n_{\text {lens }}-n_{\text {liquid }}\right)}{n_{\text {liquid }}}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Here, $n_{\{\text {lens }\}}-n_{\{\text {liquid }\}}=0$
Therefore, the focal length will be infinite.

Topic: Physics, Ray Optics, MHT CET, MHT CET 2022 (08 Aug Shift 1)
A convex lens is dipped in a liquid whose refractive index is equal to refractive index of lens material. Then its focal length will

Options:
A increase
B remain unchanged
C become infinite
D become zero

The correct answer is:
become infinite

Solution:
For a lens $\frac{1}{\mathrm{f}}=\left(\mu^{\prime}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
Where $\mu^{\prime}=\frac{\mu_{\mathrm{g}}}{\mu_{\ell}}=1$
$$
\therefore \frac{1}{\mathrm{f}}=0 \text { or } \mathrm{f}=\infty
$$

Topic: Physics, Ray Optics, MHT CET, MHT CET 2021 (21 Sep Shift 2)
A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will:

Options:
A become zero
B become infinite
C become small, but non-zero
D remain unchanged

The correct answer is:
become infinite

Solution:
When refractive index of lens is equal to the refractive index of liquid, the lens behave like a plane surface with focal length infinity.

Topic: Physics, Ray Optics, JEE Main
A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will:

Options:
A become zero
B become infinite
C become small, but non-zero
D remain unchanged

The correct answer is:
become infinite

Solution:
When refractive index of lens is equal to the refractive index of liquid, the lens behave like a plane surface with focal length infinity.

Topic: Physics, Ray Optics, NEET, NEET 2003
A convex lens is immersed in a liquid, whose refractive index is equal to the refractive index of the material of the lens. Then its focal length will

Options:
A decrease
B become zero
C become infinite
D remain the same

The correct answer is:
become infinite

Solution:
Given, aÎ¼g= aÎ¼l
The focal length of convex lens in liquid f is given by
1f= aÎ¼g aÎ¼l-11R1-1R2
1f=1-11R1-1R2
1f=0
Or f= â
Its focal length will become infinite.

Topic: Physics, Ray Optics, NEET
A convex lens is made up of three different materials as shown in the figure. For a point object placed on its axis, the number of images formed are

Options:
A 1
B 5
C 4
D 3

The correct answer is:
3

Solution:
Number of images $=$ (Number of materials)

Topic: Physics, Ray Optics, JEE Main
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in between the light source and the lens. To get the sharp image again, the screen is shifted by a distance d. Then d is:

Options:
A 0
B 0.55 cm away from the lens
C 0.55 cm towards the lens
D 1.1 cm away from the lens

The correct answer is:
0.55 cm away from the lens

Solution:
From given data
1v-1u=1f
110-1(-10)=1f
f=5cm
When glass plate is placed, shift in object for lens
s=t1-1μ=1.51-23
=0.5 cm
Shift will be in the direction of incoming rays
∴u=10-0.5=9.5 cm
1v-1u=1f
1v=15+1(-9.5)
v=47.54.5 cm
Shift =47.54.5-10=2.54.5=59cm=0.55 cm away from lens.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A convex lens of crown glass $(n=1.525)$ will behave as a divergent lens if immersed in

Options:
A Water $(n=1.33)$
B In a medium of $n=1.525$
C Carbon disulphide $n=1.66$
D It cannot act as a divergent lens

The correct answer is:
Carbon disulphide $n=1.66$

Solution:
A lens shows opposite behaviour if $\mu_{\text {medium }}\gt\mu_{\text {lens }}$

Topic: Physics, Ray Optics, JEE Main
A convex lens of focal length 10 cm and image formed by it, is at least distance of distinct vision then the magnifying power is

Options:
A 3.5
B 2.5
C 1.5
D 1.4

The correct answer is:
3.5

Solution:
The coverging lens used for magnification is called simple microscope or a magnifier.
When image is formed at D, the least distance of distinct vision, then magnifying power
M=1+Df
Given, D = 25 cm, f = 10 cm
â´M=1+2510=1+2.5=3.5

Topic: Physics, Ray Optics, JEE Main
A convex lens of focal length $0.15 \mathrm{~m}$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \mathrm{~m}$. The refractive index of the liquid is

Options:
A $\frac{7}{4}$
B $\frac{5}{4}$
C $\frac{9}{4}$
D $\frac{3}{2}$

The correct answer is:
$\frac{5}{4}$

Solution:
$$
f_a=0.15 \mathrm{~m}, \mu_g=3 / 2, f_e=0.225+0.15
$$
From lens formula,
For air, $\frac{1}{f_a}=\left(\frac{\mu_g}{\mu_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For liquid, $\frac{1}{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\therefore \begin{aligned} \frac{1 / f_a}{1 / f_l} & =\frac{\left(\mu_g / \mu_a-1\right)}{\left(\frac{\mu_g}{\mu_l}-1\right)} \\ \frac{f_l}{f_a} & =\frac{\left(\frac{3 / 2}{1}-1\right)}{\left(\frac{3 / 2}{\mu_l}-1\right)} \\ \frac{0.375}{0.15} & =\frac{1 / 2}{\left(\frac{3}{2 \mu_l}-1\right)} \\ \left(\frac{3}{2 \mu_l}-1\right) & =\frac{1 / 2}{5 / 2} \\ \frac{3}{2 \mu_l}-1 & =\frac{1}{5} \\ \frac{3}{2 \mu_l} & =\frac{1}{5}+1 \Rightarrow \frac{3}{2 \mu_l}=\frac{6}{5} \\ \mu_l & =\frac{5}{4}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
A convex lens of focal length \( 15 \mathrm{~cm} \) and a concave mirror of focal length \( 30 \mathrm{~cm} \) are kept with their optic axis PQ and RS parallel but separated in vertical direction by \( 0.6 \mathrm{~cm} \) as shown. The distance between the lens and mirror is \( 30 \mathrm{~cm} \). An upright object \( \mathrm{AB} \) of height \( 1.2 \mathrm{~cm} \) is placed on the optic axis \( \mathrm{PQ} \) of the lens at a distance of \( 20 \mathrm{~cm} \) from the lens. If \( \mathrm{A}^{\prime} \mathrm{B}^{\prime} \) is the image after refraction from the lens and the reflection from the mirror, find the distance of \( A^{\prime} B^{\prime} \) from the pole of the mirror and obtain its magnification. Also locate positions of \( A^{\prime} \) and \( B^{\prime} \) with respect to the optic axis \( R S \).

Options:
A \( -1 \cdot 8 \mathrm{~cm} \)
B \( -2.0 \mathrm{~cm} \)
C \( -18 \mathrm{~cm} \)
D \( -20 \mathrm{~cm} \)

The correct answer is:
\( -1 \cdot 8 \mathrm{~cm} \)

Solution:
Rays coming from object AB first refract from the lens and then reflect from the mirror.
Refraction from the lens
μ = - 20 cm, f = + 15 cm
Using lens formula, 1 υ - 1 u = 1 f
1 υ - 1 - 2 0 = 1 1 5
υ = + 60 cm
and linear magnification, m 1 = υ u = + 6 0 - 2 0 = - 3
i,e., first image formed by the lens will be at 60 cm from it ( or 30 cm from mirror ) towards left and 3 times magnified but inverted. Length of first image A1 B1 would be 1.2 x 3 = 3.6 cm ( inverted).

Reflections from mirror Image formed by lens ( A1 B1) will behave like a virtual object for mirror at a distance of 30 cm from it as shown. Therefore u =+ 30 cm, f =- 30 cm.
Using mirror formula, 1 υ + 1 u = 1 f
or, 1 υ + 1 3 0 = - 1 3 0
∴ υ = - 1 5 cm
and linear magnification, m 2 = - υ u = - - 1 5 + 3 0 = + 1 2
i.e., final image A' B' will be located at a distance of 15 cm from the mirror ( towards right) and since magnification is + 1 2 , length of final image would be 3 · 6 × 1 2 = 1 · 8 cm.
∴ A ′ B ′ = 1 · 8 cm
Point B is 0.6 cm above the optic axis of mirror, therefore, its image B' would be 0 · 6 ( 1 2 ) = 0 · 3 cm. above optic axis. Similarly, point A is 3 cm below the optic axis, therefore, its image A' will be 3 × 1 2 = 1 · 5 cm. below the optic axis as shown below
Total magnification of the image,
m = m 1 × m 2 = - 3 + 1 2 = - 3 2
∴ A ′ B ′ = m AB = - 3 2 1 · 2 = - 1 · 8 cm
Note that, there is no need of drawing the ray diagram if not asked in the question. Note With reference to the pole an optical instrument ( whether it is a lens or a mirror) the coordinates of the object (X0, Y0) are generally known to us. The corresponding coordinates of image ( Xi, Yi) are found as follows
X 1 is obtained using 1 υ ± 1 u = 1 f
Here, υ is actually X0 and u is X ie, the above formula can
be written as 1 X i ± 1 X 0 = 1 f
Similarly, Y is obtained from m = I 0
Here, I is Y and O is Y i.e., the above formula can be written as m = Y i / Y 0 or Y i = m Y 0

Topic: Physics, Ray Optics, JEE Main
A convex lens of focal length \( 15 \mathrm{~cm} \) is placed on a plane mirror. An object is placed at \( 30 \mathrm{~cm} \) from the lens. The image
formed is:

Options:
A Real, at \( 30 \mathrm{~cm} \) in front of the mirror
B Real, at \( 30 \mathrm{~cm} \) behind the mirror
C Real, at \( 10 \mathrm{~cm} \) in front of the mirror
D Virtual, at \( 10 \mathrm{~cm} \) behind the mirror

The correct answer is:
Real, at \( 10 \mathrm{~cm} \) in front of the mirror

Solution:
1F=1f1+1fm+1f1=2f1+1∞⇒F=152 cm
215=130+1v,v=10 cm
∴ image is real, 10 cm in front of mirror.

Topic: Physics, Ray Optics, JEE Main
A convex lens (of focal length 20 cm ) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be:

Options:
A 30 cm
B 25 cm
C 20 cm
D 10 cm

The correct answer is:
10 cm

Solution:
Image formed by lens
1v-1u=1f
⇒1v+130=120
⇒1v=120-130
⇒v=60 cm
In the image remains at the same position even if the concave mirror is removed indicates that the image formed by lens is formed at the center of curvature of spherical mirror.
∴ Radius of curvature of mirror =80-60=20 cm
Focal length, f=10 cm
For virtual image to be formed, object is to be kept between focus and pole. Hence, the maximum distance of object from spherical mirror for which virtual image is formed will be its focal length 10cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances x1 and x2x1>x2 from the lens. The ratio of x1 and x2 is:

Options:
A 2:1
B 5:3
C 4:3
D 3:1

The correct answer is:
3:1

Solution:
Given, m=2⇒m=±2
From this lens formula, 1v+1u=1f
uv+1=uf
1m+1=uf
u=f1m+1
x1=201+12
x2=201-12
x1x2=31

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A convex lens of focal length 20 cm is placed in front of convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is _____ cm.

Options:

The correct answer is:
10

Solution:
For lens 1v-1u=1f

⇒1v-1-60=120⇒1v+160=120

v=30 cm

For the final image to be formed on the object itself, after refraction from lens the ray should meet the mirror perpendicularly and the image by lens should be on the centre of curvature of mirror.

R=30-10=20 cm.

Now the focal length of mirror fm=R2=10 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and a plane mirror are arranged as shown. For an object kept at a distance of 60 cm from the convex lens, the final image, formed by the combination, is a real image, at a distance of:

Options:
A 60 cm from the convex lens
B 60 cm form the concave lens
C 70 cm from the convex lens
D 70 cm from the concave lens

The correct answer is:
60 cm from the convex lens

Solution:
From lens formula for convex lens, we have
1f=1v-1u
⇒130=1v+160
⇒160=1v
Applying lens formula to concave lens, we have
1f=1v-1u
⇒1-120=1v-140, ∵u=+40 cm
⇒v=60 cm
Virtual object 10 cm behind plane mirror.
Hence real image 10 cm in front of mirror.
Hence 60 cm from convex lens.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2016 (09 Apr Online)
A convex lens of focal length 40 cm forms an image of an extended source of light on a photoelectric cell. A current I is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm. The photoelectric current now is

Options:
A I2
B 4I
C 2I
D I

The correct answer is:
I

Solution:
For a light of any given frequency, the photoelectric current is directly proportional to the intensity of light.

As the diameter of the lens remains the same, both lenses will collect the same amount of light. Therefore, the intensity of the light will remain the same.

Hence, this is the correct option.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A convex lens of refractive index 1.5 and focal length 18 cm in air is immersed in water. The change in focal length of the lens will be_____cm.

(Given refractive index of water=43)

Options:

The correct answer is:
54

Solution:
From Lens Maker's formula,
For air: 1fa=μgμa-1 1R1-1R2
For water: 1fw=μgμw-1 1R1-1R2
∴ 1fa1fw=μgμa-1μgμw-1

⇒fwfa=1.5-11.5×34-1⇒fw=0.50.125×18=72 cm

Thus, change in focal length of the lens is 72-18=54 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (24 Jan Shift 2)