In a double-slit experiment, what is the linear distance on the screen between adjacent maxima if the wavelength is \(546 \mathrm{nm}\), the slit separation is \(0.100 \mathrm{mm},\) and the slit-screen separation is \(20.0 \mathrm{cm} ?\)

Short Answer

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Question: Determine the linear distance between adjacent maxima on the screen in a double-slit experiment, given a wavelength of 546 nm, a slit separation of 0.100 mm, and a slit-screen separation of 20.0 cm. Answer: The linear distance between adjacent maxima in this double-slit experiment is 2.184 mm.

Step by step solution

01

First, we need to convert the given values into meters for a consistent unit. The wavelength is given in \(nm\), slit separation in \(mm\), and slit-screen separation is given in \(cm\), so let's convert them: Wavelength (\(\lambda\)) = \(546 \times 10^{-9}\) meters Slit separation (\(d\)) = \(0.100 \times 10^{-3}\) meters Slit-screen separation (\(L\)) = \(20.0 \times 10^{-2}\) meters #Step 2: Apply the formula for maxima positions#

Next, we apply the formula \(y = m\frac{L\lambda}{d}\) to find the positions of two consecutive maxima, for example \(m = 1\) and \(m = 2\). For \(m = 1\): \(y_1 = 1 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) For \(m = 2\): \(y_2 = 2 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) #Step 3: Calculate the linear distance between adjacent maxima#
02

To find the linear distance between adjacent maxima, we need to find the difference between \(y_1\) and \(y_2\). Linear Distance (\(\Delta y\)) = \(y_2 - y_1\) Substitute the expressions for \(y_1\) and \(y_2\): \(\Delta y = 2 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})} - 1 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) Simplify the expression: \(\Delta y = \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) #Step 4: Find the numerical value of the linear distance#

Now, we can calculate the linear distance by substituting the values and evaluating the expression: \(\Delta y = \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})} = 0.002184\) meters Finally, convert the result to millimeters for better readability: Linear Distance (\(\Delta y\)) = \(2.184\) mm

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