282391
A convex lens in made of 3 layers of glass of 3 different material as in the figure. A point object is placed on its axis. The number of images of the object are
1 3
2 4
3 1
4 2
Explanation:
C: A lens made up of three different materials as shown have only one focal length. Thus, for a given object there is only one image.
Lens formula- \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
COMEDK 2020
Ray Optics
282392
A lens \((n=1.5)\) is placed in a liquid. To make it disappear, the value of \(n\) of liquid should be
1 \(\mathrm{n}<1.5\)
2 \(\mathrm{n}=1.5\)
3 \(\mathrm{n}>1.5\)
4 any \(\mathrm{n}\)
Explanation:
B: Given,
Refractive index of lens \((\mu)=1.5\)
To make the lens disappear, the value of refractive index of liquid should be equal to refractive index of lens.
i.e., 1.5.
GUJCET 2020
Ray Optics
282393
For glass lens \(f=+50 \mathrm{~cm}\). Then power of lens is
1 \(+0.02 \mathrm{D}\)
2 \(-2 \mathrm{D}\)
3 \(+2 \mathrm{D}\)
4 \(0.02 \mathrm{D}\)
Explanation:
C Given,
\(\mathrm{f}=50 \mathrm{~cm}\)
Power of lens, \(P=\frac{100}{\mathrm{f}}=\frac{100}{50}\)
\(\mathrm{P}=2 \mathrm{D}\)
GUJCET 2020
Ray Optics
282394
A small object is placed in the air, at a distance \(45 \mathrm{~cm}\) from a convex refracting surface of radius of curvature \(15 \mathrm{~cm}\). If the surface separates air from glass of refractive index 1.5, then the position of image is
1 \(100 \mathrm{~cm}\)
2 \(120 \mathrm{~cm}\)
3 \(125 \mathrm{~cm}\)
4 \(135 \mathrm{~cm}\)
Explanation:
D: Given,
Refractive index of air \(\left(\mu_1\right)=1\)
Refractive index of glass \(\left(\mu_2\right)=1.5\)
\(\mu_1=1, \mu_2=1.5, \mathrm{u}=-45 \mathrm{~cm}, \mathrm{R}=15 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\
\frac{1.5}{\mathrm{~V}}-\frac{1}{(-45)}=\frac{1.5-1}{15} \\
\frac{1.5}{\mathrm{~V}}=\frac{0.5}{15}-\frac{1}{45} \\
\mathrm{~V}=1.5 \times 90=135 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-14.09.2020
Ray Optics
282395
If the image of an object is at the focal point \(f\) to the right side of a convex lens, the position of the object on the left of the lens is at
282391
A convex lens in made of 3 layers of glass of 3 different material as in the figure. A point object is placed on its axis. The number of images of the object are
1 3
2 4
3 1
4 2
Explanation:
C: A lens made up of three different materials as shown have only one focal length. Thus, for a given object there is only one image.
Lens formula- \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
COMEDK 2020
Ray Optics
282392
A lens \((n=1.5)\) is placed in a liquid. To make it disappear, the value of \(n\) of liquid should be
1 \(\mathrm{n}<1.5\)
2 \(\mathrm{n}=1.5\)
3 \(\mathrm{n}>1.5\)
4 any \(\mathrm{n}\)
Explanation:
B: Given,
Refractive index of lens \((\mu)=1.5\)
To make the lens disappear, the value of refractive index of liquid should be equal to refractive index of lens.
i.e., 1.5.
GUJCET 2020
Ray Optics
282393
For glass lens \(f=+50 \mathrm{~cm}\). Then power of lens is
1 \(+0.02 \mathrm{D}\)
2 \(-2 \mathrm{D}\)
3 \(+2 \mathrm{D}\)
4 \(0.02 \mathrm{D}\)
Explanation:
C Given,
\(\mathrm{f}=50 \mathrm{~cm}\)
Power of lens, \(P=\frac{100}{\mathrm{f}}=\frac{100}{50}\)
\(\mathrm{P}=2 \mathrm{D}\)
GUJCET 2020
Ray Optics
282394
A small object is placed in the air, at a distance \(45 \mathrm{~cm}\) from a convex refracting surface of radius of curvature \(15 \mathrm{~cm}\). If the surface separates air from glass of refractive index 1.5, then the position of image is
1 \(100 \mathrm{~cm}\)
2 \(120 \mathrm{~cm}\)
3 \(125 \mathrm{~cm}\)
4 \(135 \mathrm{~cm}\)
Explanation:
D: Given,
Refractive index of air \(\left(\mu_1\right)=1\)
Refractive index of glass \(\left(\mu_2\right)=1.5\)
\(\mu_1=1, \mu_2=1.5, \mathrm{u}=-45 \mathrm{~cm}, \mathrm{R}=15 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\
\frac{1.5}{\mathrm{~V}}-\frac{1}{(-45)}=\frac{1.5-1}{15} \\
\frac{1.5}{\mathrm{~V}}=\frac{0.5}{15}-\frac{1}{45} \\
\mathrm{~V}=1.5 \times 90=135 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-14.09.2020
Ray Optics
282395
If the image of an object is at the focal point \(f\) to the right side of a convex lens, the position of the object on the left of the lens is at
282391
A convex lens in made of 3 layers of glass of 3 different material as in the figure. A point object is placed on its axis. The number of images of the object are
1 3
2 4
3 1
4 2
Explanation:
C: A lens made up of three different materials as shown have only one focal length. Thus, for a given object there is only one image.
Lens formula- \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
COMEDK 2020
Ray Optics
282392
A lens \((n=1.5)\) is placed in a liquid. To make it disappear, the value of \(n\) of liquid should be
1 \(\mathrm{n}<1.5\)
2 \(\mathrm{n}=1.5\)
3 \(\mathrm{n}>1.5\)
4 any \(\mathrm{n}\)
Explanation:
B: Given,
Refractive index of lens \((\mu)=1.5\)
To make the lens disappear, the value of refractive index of liquid should be equal to refractive index of lens.
i.e., 1.5.
GUJCET 2020
Ray Optics
282393
For glass lens \(f=+50 \mathrm{~cm}\). Then power of lens is
1 \(+0.02 \mathrm{D}\)
2 \(-2 \mathrm{D}\)
3 \(+2 \mathrm{D}\)
4 \(0.02 \mathrm{D}\)
Explanation:
C Given,
\(\mathrm{f}=50 \mathrm{~cm}\)
Power of lens, \(P=\frac{100}{\mathrm{f}}=\frac{100}{50}\)
\(\mathrm{P}=2 \mathrm{D}\)
GUJCET 2020
Ray Optics
282394
A small object is placed in the air, at a distance \(45 \mathrm{~cm}\) from a convex refracting surface of radius of curvature \(15 \mathrm{~cm}\). If the surface separates air from glass of refractive index 1.5, then the position of image is
1 \(100 \mathrm{~cm}\)
2 \(120 \mathrm{~cm}\)
3 \(125 \mathrm{~cm}\)
4 \(135 \mathrm{~cm}\)
Explanation:
D: Given,
Refractive index of air \(\left(\mu_1\right)=1\)
Refractive index of glass \(\left(\mu_2\right)=1.5\)
\(\mu_1=1, \mu_2=1.5, \mathrm{u}=-45 \mathrm{~cm}, \mathrm{R}=15 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\
\frac{1.5}{\mathrm{~V}}-\frac{1}{(-45)}=\frac{1.5-1}{15} \\
\frac{1.5}{\mathrm{~V}}=\frac{0.5}{15}-\frac{1}{45} \\
\mathrm{~V}=1.5 \times 90=135 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-14.09.2020
Ray Optics
282395
If the image of an object is at the focal point \(f\) to the right side of a convex lens, the position of the object on the left of the lens is at
282391
A convex lens in made of 3 layers of glass of 3 different material as in the figure. A point object is placed on its axis. The number of images of the object are
1 3
2 4
3 1
4 2
Explanation:
C: A lens made up of three different materials as shown have only one focal length. Thus, for a given object there is only one image.
Lens formula- \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
COMEDK 2020
Ray Optics
282392
A lens \((n=1.5)\) is placed in a liquid. To make it disappear, the value of \(n\) of liquid should be
1 \(\mathrm{n}<1.5\)
2 \(\mathrm{n}=1.5\)
3 \(\mathrm{n}>1.5\)
4 any \(\mathrm{n}\)
Explanation:
B: Given,
Refractive index of lens \((\mu)=1.5\)
To make the lens disappear, the value of refractive index of liquid should be equal to refractive index of lens.
i.e., 1.5.
GUJCET 2020
Ray Optics
282393
For glass lens \(f=+50 \mathrm{~cm}\). Then power of lens is
1 \(+0.02 \mathrm{D}\)
2 \(-2 \mathrm{D}\)
3 \(+2 \mathrm{D}\)
4 \(0.02 \mathrm{D}\)
Explanation:
C Given,
\(\mathrm{f}=50 \mathrm{~cm}\)
Power of lens, \(P=\frac{100}{\mathrm{f}}=\frac{100}{50}\)
\(\mathrm{P}=2 \mathrm{D}\)
GUJCET 2020
Ray Optics
282394
A small object is placed in the air, at a distance \(45 \mathrm{~cm}\) from a convex refracting surface of radius of curvature \(15 \mathrm{~cm}\). If the surface separates air from glass of refractive index 1.5, then the position of image is
1 \(100 \mathrm{~cm}\)
2 \(120 \mathrm{~cm}\)
3 \(125 \mathrm{~cm}\)
4 \(135 \mathrm{~cm}\)
Explanation:
D: Given,
Refractive index of air \(\left(\mu_1\right)=1\)
Refractive index of glass \(\left(\mu_2\right)=1.5\)
\(\mu_1=1, \mu_2=1.5, \mathrm{u}=-45 \mathrm{~cm}, \mathrm{R}=15 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\
\frac{1.5}{\mathrm{~V}}-\frac{1}{(-45)}=\frac{1.5-1}{15} \\
\frac{1.5}{\mathrm{~V}}=\frac{0.5}{15}-\frac{1}{45} \\
\mathrm{~V}=1.5 \times 90=135 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-14.09.2020
Ray Optics
282395
If the image of an object is at the focal point \(f\) to the right side of a convex lens, the position of the object on the left of the lens is at
282391
A convex lens in made of 3 layers of glass of 3 different material as in the figure. A point object is placed on its axis. The number of images of the object are
1 3
2 4
3 1
4 2
Explanation:
C: A lens made up of three different materials as shown have only one focal length. Thus, for a given object there is only one image.
Lens formula- \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
COMEDK 2020
Ray Optics
282392
A lens \((n=1.5)\) is placed in a liquid. To make it disappear, the value of \(n\) of liquid should be
1 \(\mathrm{n}<1.5\)
2 \(\mathrm{n}=1.5\)
3 \(\mathrm{n}>1.5\)
4 any \(\mathrm{n}\)
Explanation:
B: Given,
Refractive index of lens \((\mu)=1.5\)
To make the lens disappear, the value of refractive index of liquid should be equal to refractive index of lens.
i.e., 1.5.
GUJCET 2020
Ray Optics
282393
For glass lens \(f=+50 \mathrm{~cm}\). Then power of lens is
1 \(+0.02 \mathrm{D}\)
2 \(-2 \mathrm{D}\)
3 \(+2 \mathrm{D}\)
4 \(0.02 \mathrm{D}\)
Explanation:
C Given,
\(\mathrm{f}=50 \mathrm{~cm}\)
Power of lens, \(P=\frac{100}{\mathrm{f}}=\frac{100}{50}\)
\(\mathrm{P}=2 \mathrm{D}\)
GUJCET 2020
Ray Optics
282394
A small object is placed in the air, at a distance \(45 \mathrm{~cm}\) from a convex refracting surface of radius of curvature \(15 \mathrm{~cm}\). If the surface separates air from glass of refractive index 1.5, then the position of image is
1 \(100 \mathrm{~cm}\)
2 \(120 \mathrm{~cm}\)
3 \(125 \mathrm{~cm}\)
4 \(135 \mathrm{~cm}\)
Explanation:
D: Given,
Refractive index of air \(\left(\mu_1\right)=1\)
Refractive index of glass \(\left(\mu_2\right)=1.5\)
\(\mu_1=1, \mu_2=1.5, \mathrm{u}=-45 \mathrm{~cm}, \mathrm{R}=15 \mathrm{~cm}\)
Using formula -
\(\begin{aligned}
\frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\
\frac{1.5}{\mathrm{~V}}-\frac{1}{(-45)}=\frac{1.5-1}{15} \\
\frac{1.5}{\mathrm{~V}}=\frac{0.5}{15}-\frac{1}{45} \\
\mathrm{~V}=1.5 \times 90=135 \mathrm{~cm}
\end{aligned}\)
TS-EAMCET-14.09.2020
Ray Optics
282395
If the image of an object is at the focal point \(f\) to the right side of a convex lens, the position of the object on the left of the lens is at
