Question 1

A thin double-convex lens is intended to focus the image of an object onto a screen so that the image is life-sized on the screen. The lens has equal-magnitude radii of $83 \mathrm {~cm}$ and a refractive index of 1.6.
(a) What is the distance $d$ from the image to the screen?
(b) What is the total distance $L$ between the object and the image?

Accepted Answer

For a life-sized image, the magnification (M) is 1. The lens maker's formula, $1/f = (n-1)(1/R_1 - 1/R_2)$, where $n$ is the refractive index and $R_1$ and $R_2$ are the radii of curvature, is used to find the focal length $f$. Given $R_1 = 83 \, \text{cm}$, $R_2 = -83 \, \text{cm}$ (since it's double-convex and the radii have opposite signs), and $n = 1.6$, we find $f = 69 \, \text{cm}$. For a life-sized image, the object and image distances from the lens must both be twice the focal length, so $d = 2f = 138 \, \text{cm}$, and the total distance $L = 2d = 276 \, \text{cm}$.

Question 2

A double-convex thin glass lens has equal-magnitude radii of curvature. The focal length of the Iens is $+ 37.3 \mathrm {~cm}$ and the index of refraction of the glass is $1.52$. What is the magnitude of the radius of curvature of each convex surface?

Accepted Answer

Using the lensmaker's equation: $ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $, and given $ R_1 = R_2 = R $, we solve for $ R $ using $ f = 37.3 \, \text{cm} $ and $ n = 1.52 $. This results in $ R \approx 39 \, \text{cm} $.

Question 3

An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of magnitudes 40 cm and 60 cm. The glass has an index of refraction of 1.54. What is the focal length
Of this lens when used in the night air?

Accepted Answer

The answer of An amateur astronomer grinds a double convex...

Question 4

A double-convex thin glass lens has equal-magnitude radii of curvature. The focal length of this lens is $+ 31.2 \mathrm {~cm}$ and the index of refraction of the glass is $1.52$. The lens is to be replaced with a plano-convex glass lens of the same focal length and thickness. The radius of curvature of the convex surface of the replacement lens is $13.0 \mathrm {~cm}$. What should be the index of refraction of the glass of the plano-convex lens?

Accepted Answer

The focal length $f$ of a lens in air is given by the lensmaker's formula: $$ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$ where $n$ is the index of refraction of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the two lens surfaces. For a double-convex lens with equal radii of curvature, $R_1 = -R_2$, and for a plano-convex lens, $R_2 = \infty$.Given that the double-convex lens and the plano-convex lens have the same focal length, we can set their $1/f$ values equal to each other to find the index of refraction for the plano-convex lens.For the double-convex lens, $n = 1.52$ and $f = +31.2$ cm. For the plano-convex lens, $R_1 = 13.0$ cm and $R_2 = \infty$, so $\frac{1}{R_2} = 0$.Setting the two $1/f$ values equal and solving for the index of refraction of the plano-convex lens gives us $n = 1.42$.

Question 5

The radius of curvature of the curved side of a plano-convex lens made of glass, having index of refraction 1.64, is 33 cm. What is the focal length of the lens?

Accepted Answer

The focal length $ f $ of a plano-convex lens can be calculated using the lens maker's formula:  $$ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$  For a plano-convex lens, $ R_2 = \infty $ and $ R_1 = 33 \, \text{cm} $.  Substituting the values:  $$ \frac{1}{f} = (1.64 - 1) \left( \frac{1}{33} - 0 \right) $$  $$ \frac{1}{f} = 0.64 \times \frac{1}{33} $$  $$ f = \frac{33}{0.64} \approx 51.56 \, \text{cm} $$  Thus, the focal length is approximately 52 cm.

Question 6

A double-concave lens has equal radii of curvature of magnitudes 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.00 cm from the lens. What is the index of refraction
Of the lens material?

Accepted Answer

The correct answer is A.The lens equation is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. For a double-concave lens, f is negative. We can use the thin lens equation to find the focal length:1/f = 1/do + 1/di1/f = 1/14.2 cm + 1/(-5.00 cm)f = -7.10 cmThe radius of curvature is given by R = 2f, so R = -14.2 cm. The index of refraction is given by n = 1 + (R/f), so n = 1 + (-14.2 cm/-7.10 cm) = 1.98.

Question 7

A plano-convex lens is to have a focal length of magnitude 40 cm, and it is made of glass of index of refraction 1.65. What radius of curvature is required?

Accepted Answer

Using the lensmaker's equation for a plano-convex lens, $ \frac{1}{f} = (n - 1) \left( \frac{1}{R} \right) $, where $ f = 40 $ cm and $ n = 1.65 $, solving for $ R $ gives $ R = 26 $ cm.

Quiz 22: Electromagnetic Waves

A double-convex thin glass lens has equal-magnitude radii of curvature. The focal length of the Iens is $+ 37.3 \mathrm {~cm}$ and the index of refraction of the glass is $1.52$ . What is the magnitude of the radius of curvature of each convex surface?

An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of magnitudes 40 cm and 60 cm. The glass has an index of refraction of 1.54. What is the focal length Of this lens when used in the night air?

The radius of curvature of the curved side of a plano-convex lens made of glass, having index of refraction 1.64, is 33 cm. What is the focal length of the lens?

A double-concave lens has equal radii of curvature of magnitudes 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.00 cm from the lens. What is the index of refraction Of the lens material?

A plano-convex lens is to have a focal length of magnitude 40 cm, and it is made of glass of index of refraction 1.65. What radius of curvature is required?

Quiz 22: Electromagnetic Waves

A double-convex thin glass lens has equal-magnitude radii of curvature. The focal length of the Iens is +37.3 cm+ 37.3 \mathrm {~cm}+37.3 cm and the index of refraction of the glass is 1.521.521.52 . What is the magnitude of the radius of curvature of each convex surface?

An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of magnitudes 40 cm and 60 cm. The glass has an index of refraction of 1.54. What is the focal length Of this lens when used in the night air?

The radius of curvature of the curved side of a plano-convex lens made of glass, having index of refraction 1.64, is 33 cm. What is the focal length of the lens?

A double-concave lens has equal radii of curvature of magnitudes 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.00 cm from the lens. What is the index of refraction Of the lens material?

A plano-convex lens is to have a focal length of magnitude 40 cm, and it is made of glass of index of refraction 1.65. What radius of curvature is required?

A double-convex thin glass lens has equal-magnitude radii of curvature. The focal length of the Iens is $+ 37.3 \mathrm {~cm}$ and the index of refraction of the glass is $1.52$ . What is the magnitude of the radius of curvature of each convex surface?