Edits to “Statement”, “Solution”, “Answer”
en/13.2.15.md
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| ### Statement | |||
| − | $13.2.15.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $13.2.15.$ | |||
| a. Find the focal length of a biconvex lens with a radius of curvature of $30 \text{ cm}$, made of glass with a refractive index of $1.6$. What is the optical power of the lens? | |||
| b. One surface of a lens made of glass with a refractive index of $1.6$ is flat, the other is spherical. The optical power of the lens is $1 \text{ D}$. Determine the radius of curvature of the spherical surface of the lens. | |||
| ### Solution | |||
| The optical power of a thin lens $D$ and its focal length $F$ are calculated using the universal formula: | |||
| $$D = \frac{1}{F} = \left(\frac{n_{\text{l}}}{n_{\text{m}}} - 1\right) \left( \pm \frac{1}{R_1} \pm \frac{1}{R_2} \right)$$ | |||
| where $n_{\text{l}}$ is the refractive index of the lens material, and $n_{\text{m}}$ is the refractive index of the surrounding medium. | |||
| The "$+$" sign in the brackets is used if the surface is convex, and the "$-$" sign if the surface is concave. The values of the radii of curvature $R_1$ and $R_2$ are substituted by their absolute values. | |||
| Since the medium is not specified in the problem, we assume the lens is in air ($n_{\text{m}} = 1$). The refractive index of glass is $n_{\text{l}} = 1.6$. | |||
| @@ -21,7 +15,7 @@Solution | |||
| <b>Part a)</b> | |||
| − |  | ||
| The lens is biconvex, so both of its surfaces are convex. We place a "$+$" sign before both terms with radii. According to the problem, $R_1 = R_2 = R = 0.3 \text{ m}$. | |||
| Optical power of the lens: | |||
| $$D = \left(\frac{n_{\text{l}}}{1} - 1\right) \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2(n_{\text{l}} - 1)}{R}$$ | |||
| Substituting the values: | |||
| $$D = \frac{2(1.6 - 1)}{0.3} = \frac{1.2}{0.3} = 4 \text{ D}$$ | |||
| Focal length: | |||
| $$F = \frac{1}{D} = \frac{1}{4} = 0.25 \text{ m} = 25 \text{ cm}$$ | |||
| @@ -33,7 +27,8 @@Solution | |||
| <b>Part b)</b> | |||
| − |  | ||
| + | |||
| The lens is plano-spherical. Since the optical power of the lens is positive ($D = 1 \text{ D} > 0$), the lens is converging. It follows that its spherical surface must be <b>convex</b> (if it were concave, the lens would be diverging). | |||
| For a flat surface, the radius of curvature is $R_1 \to \infty$, so $\frac{1}{R_1} = 0$. | |||
| Taking into account the "$+$" sign for the second, convex surface, the lens formula takes the form: | |||
| $$D = (n_{\text{l}} - 1) \left( 0 + \frac{1}{R_2} \right) = \frac{n_{\text{l}} - 1}{R_2}$$ | |||
| Expressing the required radius of curvature $R_2$: | |||
| $$R_2 = \frac{n_{\text{l}} - 1}{D}$$ | |||
| Substituting the values: | |||
| $$R_2 = \frac{1.6 - 1}{1} = 0.6 \text{ m} = 60 \text{ cm}$$ | |||
| @@ -46,7 +41,3 @@Solution | |||
| #### Answer | |||
| a. $F = 0.25 \text{ m}$; $D = 4 \text{ D}$ | |||
| b. $R = 0.6 \text{ m}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $13.2.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $13.2.15.$ | $13.2.15.$ | ||
| a. Find the focal length of a biconvex lens with a radius of curvature of $30 \text{ cm}$, made of glass with a refractive index of $1.6$. What is the optical power of the lens? | a. Find the focal length of a biconvex lens with a radius of curvature of $30 \text{ cm}$, made of glass with a refractive index of $1.6$. What is the optical power of the lens? | ||
| b. One surface of a lens made of glass with a refractive index of $1.6$ is flat, the other is spherical. The optical power of the lens is $1 \text{ D}$. Determine the radius of curvature of the spherical surface of the lens. | b. One surface of a lens made of glass with a refractive index of $1.6$ is flat, the other is spherical. The optical power of the lens is $1 \text{ D}$. Determine the radius of curvature of the spherical surface of the lens. | ||
| ### Solution | ### Solution | ||
| The optical power of a thin lens $D$ and its focal length $F$ are calculated using the universal formula: | The optical power of a thin lens $D$ and its focal length $F$ are calculated using the universal formula: | ||
| $$D = \frac{1}{F} = \left(\frac{n_{\text{l}}}{n_{\text{m}}} - 1\right) \left( \pm \frac{1}{R_1} \pm \frac{1}{R_2} \right)$$ | $$D = \frac{1}{F} = \left(\frac{n_{\text{l}}}{n_{\text{m}}} - 1\right) \left( \pm \frac{1}{R_1} \pm \frac{1}{R_2} \right)$$ | ||
| where $n_{\text{l}}$ is the refractive index of the lens material, and $n_{\text{m}}$ is the refractive index of the surrounding medium. | where $n_{\text{l}}$ is the refractive index of the lens material, and $n_{\text{m}}$ is the refractive index of the surrounding medium. | ||
| The "$+$" sign in the brackets is used if the surface is convex, and the "$-$" sign if the surface is concave. The values of the radii of curvature $R_1$ and $R_2$ are substituted by their absolute values. | The "$+$" sign in the brackets is used if the surface is convex, and the "$-$" sign if the surface is concave. The values of the radii of curvature $R_1$ and $R_2$ are substituted by their absolute values. | ||
| Since the medium is not specified in the problem, we assume the lens is in air ($n_{\text{m}} = 1$). The refractive index of glass is $n_{\text{l}} = 1.6$. | Since the medium is not specified in the problem, we assume the lens is in air ($n_{\text{m}} = 1$). The refractive index of glass is $n_{\text{l}} = 1.6$. | ||
| @@ -21,7 +15,7 @@Solution | |||
| <b>Part a)</b> | <b>Part a)</b> | ||
|  | ||
| The lens is biconvex, so both of its surfaces are convex. We place a "$+$" sign before both terms with radii. According to the problem, $R_1 = R_2 = R = 0.3 \text{ m}$. | The lens is biconvex, so both of its surfaces are convex. We place a "$+$" sign before both terms with radii. According to the problem, $R_1 = R_2 = R = 0.3 \text{ m}$. | ||
| Optical power of the lens: | Optical power of the lens: | ||
| $$D = \left(\frac{n_{\text{l}}}{1} - 1\right) \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2(n_{\text{l}} - 1)}{R}$$ | $$D = \left(\frac{n_{\text{l}}}{1} - 1\right) \left( \frac{1}{R} + \frac{1}{R} \right) = \frac{2(n_{\text{l}} - 1)}{R}$$ | ||
| Substituting the values: | Substituting the values: | ||
| $$D = \frac{2(1.6 - 1)}{0.3} = \frac{1.2}{0.3} = 4 \text{ D}$$ | $$D = \frac{2(1.6 - 1)}{0.3} = \frac{1.2}{0.3} = 4 \text{ D}$$ | ||
| Focal length: | Focal length: | ||
| $$F = \frac{1}{D} = \frac{1}{4} = 0.25 \text{ m} = 25 \text{ cm}$$ | $$F = \frac{1}{D} = \frac{1}{4} = 0.25 \text{ m} = 25 \text{ cm}$$ | ||
| @@ -33,7 +27,8 @@Solution | |||
| <b>Part b)</b> | <b>Part b)</b> | ||
|  | ||
| The lens is plano-spherical. Since the optical power of the lens is positive ($D = 1 \text{ D} > 0$), the lens is converging. It follows that its spherical surface must be <b>convex</b> (if it were concave, the lens would be diverging). | The lens is plano-spherical. Since the optical power of the lens is positive ($D = 1 \text{ D} > 0$), the lens is converging. It follows that its spherical surface must be <b>convex</b> (if it were concave, the lens would be diverging). | ||
| For a flat surface, the radius of curvature is $R_1 \to \infty$, so $\frac{1}{R_1} = 0$. | For a flat surface, the radius of curvature is $R_1 \to \infty$, so $\frac{1}{R_1} = 0$. | ||
| Taking into account the "$+$" sign for the second, convex surface, the lens formula takes the form: | Taking into account the "$+$" sign for the second, convex surface, the lens formula takes the form: | ||
| $$D = (n_{\text{l}} - 1) \left( 0 + \frac{1}{R_2} \right) = \frac{n_{\text{l}} - 1}{R_2}$$ | $$D = (n_{\text{l}} - 1) \left( 0 + \frac{1}{R_2} \right) = \frac{n_{\text{l}} - 1}{R_2}$$ | ||
| Expressing the required radius of curvature $R_2$: | Expressing the required radius of curvature $R_2$: | ||
| $$R_2 = \frac{n_{\text{l}} - 1}{D}$$ | $$R_2 = \frac{n_{\text{l}} - 1}{D}$$ | ||
| Substituting the values: | Substituting the values: | ||
| $$R_2 = \frac{1.6 - 1}{1} = 0.6 \text{ m} = 60 \text{ cm}$$ | $$R_2 = \frac{1.6 - 1}{1} = 0.6 \text{ m} = 60 \text{ cm}$$ | ||
| @@ -46,7 +41,3 @@Solution | |||
| #### Answer | #### Answer | ||
| a. $F = 0.25 \text{ m}$; $D = 4 \text{ D}$ | a. $F = 0.25 \text{ m}$; $D = 4 \text{ D}$ | ||
| b. $R = 0.6 \text{ m}$ | b. $R = 0.6 \text{ m}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||